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8 Tauberian Theorems
8
Tauberian Theorems
1. Introduction
We recall that a series
is sermmable (A), in the sense of Abel, to the number A if the power series m
f(x)=
k= 0
akXk
converges for 1x1 < 1 and iff(1 -) = A . The classical theorem of Abel is a regularity or consistency result to the effect that the convergence of (1.1) to A implies its Abel summability to the same value. The example ak = ( - l)k shows that the converse statement is not true. It is this very fact that makes summability of divergent series a useful concept. A " corrected " converse is, however, true. For example, summability (A) implies convergence if one of the following conditions holds : 193
194
8. Tauberian Theorems
The first of these results is trivial as a consequence of the following inequalities : n
xakXksf(X),
O
k=O
n = O , 1 , 2 , ...,
Since the partial sums s, E t, they must approach a limit if f(1 -) is finite; hence summability implies convergence. The case B was first proved by A. Tauber [1897], case C by J. E. Littlewood [1910]. The latter result and closely related ones might be said to form the underlying theme of the present chapter. We recall further that series (1.1) is summable (C), in the sense of Cesaro, to the number A if lim
SO
n-t OCI
+ s1 + ..*+ s, n+l
= A.
This method of summability is also regular. (See Exercise 1 in this chapter.) The same example used above shows that summability (C) does not imply convergence. Again a “corrected” converse can be proved. For example, G . H. Hardy [1910, p. 3011 proved that condition C above is a suitable correction. We shall prove this later. Note that f(x)=
so + S I X
+ s2x2 + ...
1 S X +x2
+
...
so thatf(x) may be regarded as a weighted average, with weights xk, of all sk . Abel summability results in the limiting case when all weights are unity. Similarly, equation (1.3) asserts, in a sense, that the arithmetic average of “all” s k is A. These two examples are typical of
8.1. Introduction
195
summability theory. In general we say that a theorem is Abeliun if it asserts something about an average of a sequence from a hypothesis about its ordinary limit; it is Tuuberiun if conversely the implication goes from average to limit. Usually the latter results must have in the hypothesis some supplementary condition, like A, B, or C above. In this chapter we shall be dealing chiefly with Tauberian theorems involving integrals, which are the continuous analogs of series. After such results are proved it is then a fairly simple matter to obtain, as special cases, the classical ones about series. See $7. We conclude this section with the proof of a typical Abelian result, which will be useful later. Theorem 2.1.
Throughout this chapter we shall assume without further statement that u(t) E Lin (0, R) for every R > 0. Since
it is no restriction to assume that A = 0, u(t) = o(tY).Then to an arbitrary E > 0 there corresponds an R such that lu(t)l
Hence
< EtY,
R 5 t < co.
196
8. Tauberian Theorems
It follows that x -+ Of,
f ( x ) = o(x-y-'),
as we wished to prove. 2. Integral Analogs
In this chapter we shall be primarily concerned with Tauberian theorems involving Laplace integrals, and these may be regarded as continuous analogs of Tauber's original series theorem, or generalizations thereof. We say that the integral
is summable (A) to the value A if the Laplace integral f(x)=
(2.2)
I
m
0
e-%(t) dt
converges for x > 0 and iff(l-) = A . It is summable (C) if (2.3)
A(x) = Joxa(f) d t ,
lim
Our first result is the regularity theorem.
Theorem 2.1. 1. f ( x ) = 2. =>
JOm
m
0
e-%(t) dt,
0 < x < co;
a(t) dt = A
f(U+)
= A.
That is, the convergence of (2.1) implies its Abel summability to the same value. This result is an immediate consequence of Theorem 4.1 of Chapter 5. A converse of this, corresponding to Condition A of $1 follows.
8.2. Integral Analogs
197
Theorem 2.2. 1. f ( x ) =
m
J'0
0 < x < 00;
e-x'a(t) dt,
2. f ( O + ) = A ; 0 < t < co
3. a(t) 2 0 ,
a(t) dt
JOm
= A.
Analogous to the corresponding proof for series given in $1, this result follows easily from the inequality
A(R) = lim x+o+
loR
e "'a(t) dt -
5 l b f(x). x-o+
The example a(t) = cos t shows that hypothesis 3 cannot be omitted. For, in this case f ( x ) = x/(x' 1), f ( O + ) = 0, but the integral (2.1) diverges. Our next result is considered Abelian though it is a comparison of two kinds of averages. It states that Ceshro summability implies Abel summability.
+
Theorem 2.3. 1. f ( x ) = (2.4)
lorn0 < x < C x t a ( t )dt,
2. /:o(t)dt - A x ,
(2.5) If a(t) is an integral
then it is the integral
A
f(x)--,
X
x+
03
x+o+
00;
198
8. Tauberian Theorems
that is summable ( C ) by hypothesis 2 and which is summable (A) by (2.5). However, the theorem as stated, in terms of a(t) rather than b(t), suggests a different interpretation. The integral (2.1) diverges to + 00 ( A > 0), but the partial integrals A(x) become infinite in the prescribed way (2.4). The conclusion is that f(x) also becomes infinite, but at a resultant rate (2.5), as x + 0 . This result is essentially contained in Theorem 1. For, an integration by parts gives
+
f(x) = x
-
low e-XtA(t) dt,
x > 0,
and (2.4) becomes A(x) Ax, X + co. The conclusion of Theorem 1, y = 1, applied to the integralf(x)/x now yields the desired result. An example to show that the theorem is not reversible is a(t) = 1
(2.6)
+ sin t + t cos t , f(x)
-
1 x
=-
A(t) = t
+ t sin t
+ (x' 2x2 +
1)2'
Here f(x) l/x, x + 0 + , but A (t)/t approaches no limit as t + co. Accordingly, to obtain a corrected converse we must expect to impose some additional condition on a(t). It turns out that boundedness is a suitable one. Note that a(t) as defined by (2.6) is not bounded.
Theorem 2.4. W
1. f(x> =
J0
e-xta(t)
dt,
2. a(t) E B (bounded), 3. f(x)--, (2.7)
=r
1:
A X
a(t) dt
o < x < co; 0 < t < 00;
x+o+
-
Ax,
x + 00.
We shall obtain this result in the next section as a special case of a much more general theorem. But let us illustrate it here by the example
8.3. A Basic Theorem
199
The Laplace transform of a(t) is (Exercise 14) (2.8)
1
1
f(x) = - tan-’ X x
7
I
-
x+o+.
2x’
Since a(t) E B we may apply the theorem to obtain (2.7) with A But this may be verified directly since Joxa(f)dt = r ( x 0
sin y
.x
sin y
Y
O
Y
- y ) -d y = x J -dy
x-rm X
a(t) dt =
I
c0
O
sin y
7I
Y
2
= 42.
+ cos x - 1,
-d y = -.
Exercise 3 shows that hypothesis 2 is essential here. 3. A Basic Tauberian Theorem
We observe that hypothesis 3 of Theorem 2.4 may be written as
if g(t) = e - t and that the conclusion (2.7) takes the same form if g(t) is replaced by h(t) = 1 on (0, l), h(t) = 0 on (1, 00). Thus Theorem 2.4 states that, under the restriction of boundedness on a(& if (3.1) holds for one “ kernel ’’ g(x) it also holds for at least one other kernel h(t). Our basic theorem will show that the conclusion will hold for many kernels h(t) even if g(t) is not e-t. The essential property of e - * for our purpose is contained in the familiar uniqueness property of the Laplace transform: the vanishing of the generating function implies that of the determining function. We dignify this property in a kernel g(t) by a notation as follows. Definition 3.1. A function g(x) E U (has the uniqueness property) on (0, co)if g(x) E L on (0, co) and if the equation Jomg(;)a(t)
dt
E
0,
0 < x < co
for a(t) E B. C implies a(t) = 0, 0 < t < co.
8. Tauberian Theorems
200
Clearly an exponential change of variables converts the integral (3.1) into a convolution, the range of integration becoming the whole real axis. Our uniqueness property takes the following form for the interval (-a, a). (-
Definition 3.2. A function g(x) E U on (- co,00) if g(x) E L on co) and if the equation
rm
00,
g(x - t) a(t) dt
= 0,
0
for a(t) E B. C implies a(t) = 0, - co < t < co. Note that if g(x) E U on (0, 03) then g(e")e" E U on (- 00, m). For example e-" E U on (0, 00) and exp(-e")e" E U on (- co, 00). On the other hand a function g(x) which is constant over a finite interval and is zero elsewhere 4 U on (- 00, 00). This can be seen'by choosing a(t) as a suitable periodic function. (See Exercise 12 of this chapter.) We now state our basic Tauberian Theorem as follows.
Theorem 3.1. 1. g(x)E
u,
2. a(x) E B ,
3. h(x) EL, 4.
Jm -m
(-a, a); (-a, co);
(-a, m);
- t)a(t)dt
3
11m
g(x - t)a(t) d t --t A +A
m
m
m
g(t) dt,
h(t) d t ,
x+m
x + m.
By considering the function a(t) - A we see that there is no loss of generality in assuming A = 0. Set G=g*a,
H=h*a.
From the assumption G(m) = 0 we wish to prove H(m) = 0. Assuming the contrary, there must exist 6 > 0 and a sequence x,,tending to co with n such that H(xJ I > 6. Now observe that G * h = H * g = g * h *a. (3.2)
I
+
This follows by Fubini's theorem in the presence of hypotheses 1, 2, 3.
8.3. A Basic Theorem
Now consider the sequence = H(x
s,(x)
201
+ x,).
We shall show that it is bounded and equicontinuous on - co < x < co. For, if M is an upper bound for I a(t) 1, then
-co < x < co,
IH(x)I 5 M I m Ih(t)l dt, -m
I s,(x) - s,(y) I 5 M J”- m 1 h(x + X, - t ) - h(y + X, - t ) [ m
dt
=~(l),
y-x40.
The latter result follows from a classical result of Lebesgue theory ; see N. Wiener [1933, p. 141. By Ascoli’s lemma (see below) we now select from the functions sn(x) a subset s,(x) which tends to a limit s(x), continuous on (- co, co). By Lebesgue’s limit theorem, applicable since H E B, g EL, we have lirn
k-ao
(3.3)
1- s,,(x
= lirn k-m
m
m
- t ) g ( t ) dt
= lim k+m
m
m
J”-ao G(x + xnk- t)h(t) dt =
H(x
1
m
W
-m
+ x,,
- t)g(t) dt
s(x - t ) g ( t ) dt = g
* s.
Here we have used equation (3.2). Since G(m) = 0, another application of the Lebesgue limit theorem shows that the limit (3.3) is zero. Since s(x) E B.C we infer from hypothesis 1 and Definition 3.2 that s(x) -= 0. But Is(0)l = lirn Is,,(O) I = lim IH(x,,,)l 2 6 > 0. k+m
k-m
The contradiction shows that H(co) = 0, as we wished to prove. For the reader’s convenience we prove the form of Ascoli’s lemma which we have used.
Lemma 3.1. 1. {s,(x)}~ is equicontinuous on - co < x < co;
2.
Is,(x)l < M ,
+ A.
lim s,(x) k+ m
B. s ( x ) E C ,
n
=
= s(x),
I , 2, ..., -co < x <
00,
some M
some integers nk , some s(x);
-w
202
8. Tauberian Theorems
Arrange all real rational numbers in a sequence xl, x 2 ,x3, . . . . From the bounded double sequence s,(x,,) we may pick, by the familiar diagonal process, a subsequence snk(xm)such that (3.4)
lim snk(x,)= s(x,),
k- m
m = 1,2, 3,
. . ..
See, for example, D. V. Widder [1946, p. 261. The function s(x) is defined by the limit (3.4) at all rational points. Now let y be irrational and let E > 0 be arbitrary. By hypothesis 1, if x , is sufficiently near to y ,
I snk(y) - S n k ( x m ) I < E , IS,,(Y)
< E,
- t9,(x,)l
1,2937 . . ., j = 1,2,3, ....
k
=
Such an x , exists since y is a limit point of rational points. By Cauchy's limit criterion equation (3.4) implies the existence of a number K such that k >K, j > KISnk(Xrn) - snl(Xrn)l < E , By the triangle inequality
I S,,(Y)
- s,,(Y>
I <3 ~ ,
k >K, j >K,
from which we may infer the existence of
(3.5)
lim
k+m
S",(Y> = 4 Y > .
It remains to show only that s(x), now defined for all x, is continuous. By the equicontinuity of the sequence snk(x)we have for a given E the existence of a number 6, independent of nk, such that
1
snk(x)
- snk(y)
Is(x) - 4 Y ) I
I < 8,
s
E,
1 x - Y I < 6, Ix - Yl < 8.
Thus the lemma is proved. We point out that Ascoli's result is usually stated for compact regions, in which case the limit (3.5) is uniform. That the approach need not be uniform on (- 00, co)may be seen by considering sn(x) = sin (x/n). Here s(x) = 0, but any inequality of the form Is,(x)I < E < 1 must fail (at x = ns1/2, for example). The previous theorem, restated for the interval (0, co) becomes the following.
8.4. Hardy and Littlewood Theorems
203
Theorem 3.2.
u,
(0, co);
2. u(x) E B,
(0, co);
3. h(x) E L ,
(0, co);
1. g ( x ) E
We may specialize this result by choosing g(x) = e - x x a - l and h(x) = xP-' on (0, l), h(x) = 0 on (1, co). The hypotheses 1 and 3 are evidently satisfied if a > 0, > 0.
Theorem 3.3
2. a(x) E B, 3. f ( x )
(0, a);
- ma) XU
'
x -+
o+
In particular if a = /? = 1, this result reduces to Theorem 2.4, the proof of which we had deferred. 4. Hardy's and Littlewood's Integral Tauberian Theorems
As we noted in $1, Hardy proved first that if ak = O(l/k), k 3 co, then the series (1.1) cannot be summable in the sense of CesAro unless it converges; Littlewood (1912, p. 434) proved the same result for Abel summability. In this section we shall prove the integral analogs of these results, deriving therefrom i n 97 the classical series result. We need first a new definition.
8. Tauberian Theorems
204
Definition 4. A function f ( x ) E SO (slowly oscillating) on (0, co) if lim M
Y ) -f(x)l
=0
when y and x + 00 in such a way that y / x + 1. For example, if f ( x ) f ( x ) E SO there. For,
I f ( Y ) - f ( x >I 5
and x l f ’ ( x ) l < M on (0, co) then
E C’
lpv) I s
log($
= o(l),
L 1 .
X
Thus sin (log x ) E SO on (0, 00). On the other hand, sin x $ SO on (0, co). (See Exercise 6 of this chapter.)
Theorem 4.1.
1. a(x) E so, -x
2. Jo a(t) dt
*
-
(0, co);
AX,
co
X +
a(co) = A .
As usual, we may assume A = 0. From hypothesis 1 we see that to an arbitrary E > 0 there correspond numbers K and S such that la(y) - a(x)I
y > K, x > K,
< E,
That is, for x - 6x < y < x
+ 6x
a(x) - E
Integration gives
[a(x) - ~ 1 2 < 6 ~
Now let x Thus
+ 00.
x(li6)
x(1-a)
lim
X+ 00
a(y) dy
< [a(x) + E ] 2Sx.
The integral (4.1) is o(x), x [a(x)
- E l 2s 2 0,
-&
- 11 < 6.
< a ( y ) < a(x) + E .
1
(4.1)
:1
l&ij
[a($
-,co, by
+ E l 2s 2 0,
5 lim a(x) 5 ihi a(x) 5 E .
The conclusion is now immediate.
hypothesis 2.
8.4. Hardy and Littlewood Theorems
205
Hardy's integral Tauberian theorem is an immediate consequence of this result.
Theorem 4.2. (0, GO);
1. xa(x) E B,
2. A ( x ) =
3. (4.2)
Jx 0
lx dt ; 0
a(?)
A ( t ) d ?- A x ,
J,
~ ( tdt)
x-+GO
= A.
Hypothesis 1 corresponds to condition C of $1 and hypothesis 3 states that the integral (4.2) is summable (C) to the value A . We apply Theorem 4.1 to the function A(x). It belongs to SO since
Here M is a bound for 1 xa(x) 1. The conclusion of Theorem 4.1 is A(m) = A , equivalent to (4.2). The proof of Littlewood's integral Tauberian theorem is slightly more complicated. Theorem 4.3.
(4.3) +-
m
jo a(t) = A .
By use of Theorem 2.4 we show first that, in the presence of hypothesis 1, the Abel summability of the integral (4.3) implies its Ceshro
206
8. Tauberian Theorems
summability, and we can then apply Theorem 4.2. An obvious integration by parts enables us to write hypothesis 2 as
lom e-"'A(t) dt - A x ,
x +0
+.
To prove that A ( x ) E B, consider the difference a(t)[1 - e-'IX]
dt -
j
m
e-'lxu(t)
dt.
X
By hypothesis I
5M
i;"
+M
dt
2M.
Since f ( x ) E B on (0, co) by hypothesis 2, the same must be true of A(x). Hence if we apply Theorem 2.4 to A(x) we find that
j;A(t) dt - A x ,
x -+ co.
That is, the integral (4.3) is summable ( C ) to A , and the convergence of (4.3) now follows from Hardy's theorem. As an illustration take a(t) = (sin t)/t. Its Laplace transform is tan-'(llx) which tends to 742 as x + O + . Obviously ta(t) E B, so that
somSF dt
= 71 -
2'
5. One-sided Tauberian Conditions
In Hardy's and Littlewood's theorems the condition u(x)x EB can be replaced by a one-sided condition in which a(x)x is bounded below (or above) only. With a view to proving this we first establish several results of J. Karamata [1931, p. 271.
8.5. One-sided Conditions
Theorem 5.1. 1. g ( x ) = O , O S x < c < l ;
2. E > 0 ,
* (5.1) (5.2)
g(x)EC,
207
c5xg1;
y>o
There exist polynomials p ( x ) , P ( x ) such that p(x) < g(x) < P(x), Jol
[log(;)]y-
05x5 17
b ( x ) - p ( x ) ] d x < E.
For an arbitrary positive number q we can determine h(x) E C on 0 5 x 5 1 such that g(x)
5 h(x),
0 5 x 2 1.
For example, if g ( c + ) > 0 (the only case needed later) we may take h(x) = g(x) except in a small interval (c - 6, c) where h(x) = g(c+)(x - c + 6)/6. The value of the integral (5.3) is less than
Hence inequality (5.3) may be satisfied by choice of 6. By the Weierstrass approximation theorem we determine a polynomial Q(x) such that Ih(x) - Q(x)I < q on 0 S x 5 1, and we define P(x) = Q(x) q. Then g(x) 5 h(x) < P(x) and
+
x [(P - Q) +
I Q - h I + ( h - g)l dx <
+ Y ~ ( Y )+ V .
Similarly, there is a polynomial p(x) such that p(x) < g(x) on (0, 1) and such that
208
8. Tauberian Theorems
+
Adding inequalities (5.4) and (5.5) and choosing g = &/(4r(y) 2), we have (5.2). Hypothesis 1 could be very considerably relaxed if desired; see D. V. Widder [1941, p. 1891. Theorem 5.2.
0 < x < co;
2. a(x) 2 0 ,
3. g(x)=O, 4. f ( x )
osx
A
y>o,
xy '
N -
g(x)EC, c l x S 1 ;
x+o+.
x+o+.
We begin by determining the polynomials p and P of Theorem 5.1, corresponding to a given E . Since a(x) 2 0 we have (5.7)
/
e-"'p(e-"')u(t) dt
m
0
e-"'g(e-"')a(t) dt
5
1
W
0
e-"'P(e-"')a(t) dt,
If we replace x by (n + 1)x in hypothesis 4 we obtain
Thus the conclusion (5.6) is valid if g(x) is replaced by X" and hence by any polynomial. From (5.7) and (5.9) we have e-'p(e-')tY-' dt
5 l b xT(y) x-o+
S fi;;; x T ( y ) / x+o+
m
e-X' g(e-"')a(t) dt 0
00
0
e-"'g(e-"')a(t)
dt
5A
/
m
0
e-'P(e-')ty-'
dt.
8.6. One-sided Littlewood Theorem
209
By (5.2) the two extremes of this inequality differ by AE.Hence the two terms in the middle must be equal to each other and to the middle term of (5.8). That is, (5.6) is established. We now specialize the function g(x) to obtain the following Tauberian theorem. Theorem 5.3.
3. f ( x )
-
A
-
xy
'
y>o,
x-bo+
Except for the present one-sided Tauberian condition 2 this result would be included in Theorem 3.3. To prove it, choose g(x) = l / x on (l/e, l ) , g(x) = 0 on (0, l/e) in Theorem 5.2. Then (5.6) becomes
which is equivalent to (5.10). 6. One-sided Version of Littlewood's Integral Theorem
To obtain a one-sided version of Theorem 4.3 we need several preliminary results, of interest in themselves. We introduce a definition. Definition 6. A function f ( x ) E SD (slowly decreasing) on (0,co) if (6.1)
lim M Y ) - f W l 2 0,
when y and x become infinite in such a way that y > x and y/x + 1 .
8. Tauberian Theorems
210
For example, any increasing function belongs to this class. Or,
f(x) E SD if xf’(x) > - M for some constant M . For,
(;),
f(Y) - A x ) > -Mlog
Y > x,
from which (6.1) is immediate. We now prove a one-sided version of Theorem 4.1.
Theorem 6.1. 1. a(x) E SD,
2.
/ox
a(t) dt
-
Ax,
x + co
a(w) = A.
=>
Assume A
(0, 00);
= 0.
For an arbitrary
> 0 we have from (6.1) that x < y < (1 3. 6)x. E
a(x) - E < a(y), (6.2) Integrating with respect to y and using hypothesis 2, we obtain
lim a(x)
7
(6.3)
x-t m
5 E.
Rewrite (6.2) as
and integrate with respect to x :
This with (6.3) shows that a(co) = 0, as we wished to prove. Corollary 6.1. 1.
xa(x) > - M ,
2.
JOm
a(x) dx
someM, O < x < co;
is summable (C)to A
6.8. One-sided Littlewood Theorem
211
This is a one-sided version of Theorem 4.2. It follows at once by observing that A@) now satisfies the conditions of Theorem 6.1. We prove next a result of E. Landau [1929,p. 581.
Theorem 6.2.
1.
osx
f(X)EC2,
2. f ( 0 ) = A ; -M 3. f”(x) > x2
*
’
someM, O < x < c o
(3
f’(x)=o - ,
x-+o+.
From Taylor’s theorem we have for 0 < 6 c I
Now let x obtain
-+
0
+ , noting that f ( x + ax) - f ( x )
-+
0 by hypothesis 2.We
-M6 M6 < lim x f ‘ ( x ) 5 G xf’(x) 5 -. 2(1 - S ) 2 =xz+ x + o + 2 Since 6 is arbitrary we obtain at once the desired conclusion. We derive next a consequence of Theorem 4.3, using a modification of the first hypothesis thereof.
Theorem 6.3. 1. B(x) =
jxta(t) dt = o(x), 0
2. f ( x ) = Jorn e-x‘a(t) dt -+ A , (6.4)
=>
jo* a(t) dt = A .
x
-+
x
co ;
-+
o+
212
8. Tauberian Theorems
This states that hypothesis 1 is an alternative condition to Littiewood's for the Abel summability of the integral (6.4) to imply its convergence. Assume A = 0. Integration by parts gives (6.5) Jme-xf 1
3 dt = B(l)e-" - x Jme-"' 9 dt + 1e -%(t) t2 t m
1
1
dt.
The determining function B(t)/t2of this first integral is U ( l / t ) by hypothesis 1. Hence we may apply Theorem 4.3. Its limit as x 0 + , from the right-hand side of (6.5) is --f
B(l) -
For, lim x
x+o+
1
1
0
a(t) dt.
f,*e - X fB(t) dt = 0
by Theorem 2.1, and
lim fpe-.i2ct, dt
=
- lim
Jol
e-*'a(t) dt
1
=
- l 0 u ( t )dt
by hypothesis 2. Thus the conclusion of Theorem 4.3 yields
jlm $?dt = B ( l ) - j a(t) dt, 1
0
=B(l)
(6.7)
+
m 1
a(t) dt.
To obtain (6.7) we have integrated integral (6.6) by parts. Equation (6.4), with A = 0, is now immediate by comparing (6.6) with (6.7). We can now obtain the desired one-sided version of Littlewood's theorem very easily.
Theorem 6.4.
2. f ( x ) =
loebxta(t)dt JOm
a(t) dt
+A ,
= A.
x + O+
8.7. Classical Series Results
2 13
From hypothesis 1 we have
By Theorem 6.2f'(x) = o(l/x), x +O+. Thus Jbme-"'[M
+ ta(t)] dt = M-f ' ( x ) X
-
M
-, X
x -+
O+
Since M + ta(t) > 0 we may apply Theorem 5.3 to this integral to obtain
Jb'
[ M + tu(t)] dt
N
Mx,
x + 00.
This implies that hypothesis 1 of Theorem 6.3 is valid. The conclusion of that theorem is the desired result here. 7. Classical Series Results
We have hitherto proved the integral analogs of the Tauberian theorems about series mention in $1. It is now possible to obtain these discrete results, not by analogy, but as direct consequence of the continuous cases. We do so in a few instances, using the more general onesided versions. First, the discrete analog of Theorem 2.4 or of Theorem 5.3 (y = 1): Theorem 7.1. (7.1)
1.
2.
m
xake-kx-tA,
k=O
n
-M
x+O+;
someM, n = 0 , 1 , 2 ,...
=>
That is, if the sequence of partial sums of a series is bounded on one side it cannot be summable in the sense of Abel unless it is also summable in the sense of Ceshro. It is obvious that for y = 1 in Theorem 5.3 the lower bound zero of hypothesis 2 may be replaced by any number -M. (See Exercise 15 of this chapter.)
214
8. Tauberian Theorems
Define a function A(x) so that A(0) = 0 and A(x) =
c a, > - M ,
[XI
k=O
x > 0.
If the sum of the series (7.1) isf(x), we have
1 e-”‘A(t) dt, 03
W
f ( x ) = Jo e-”‘dA(t) = x
0
x
> 0.
Thus by hypothesis 1 j o r n e - X t A ( t ) d t -A- ,
x-bo+.
X
The conclusion of Theorem 5.3, with x replaced by n
+ 1, is
This is the desired result. We turn next to the series form of Hardy’s theorem.
Theorem 7.2.
n
n
Observe k s t that hypothesis 2 implies that sn= o(n) and hence that an = o(n), n -+ 00. Define A(x) as in the previous theorem and apply Theorem 6.1 to it. Set m = [ y ] and n = [x], y > x. Then
8.7. Classical Series Results
215
From these inequalities it is clear that A(x) E SD on (0, co). Further, if n 1 S x < n + 2
+
Then by hypothesis 2
j; A ( t ) dt
N
Ax,
x
+
co.
The conclusion of Theorem 6.1 is A(0)=
m
Uk k= 0
= A,
as we wished to prove. The series version of Littlewood’s theorem follows. Theorem 7.3. 1.
ak
-M
someM, k = 0 , 1,2, ...;
>k + 1’
W
We apply Theorem 6.4 defining a(x) by U(x)=Uk,
ksx
k = 0 , 1 , 2 ,....
From hypothesis 1 it is clear that a(x) > - M / x for 1 4 x < co. The conclusion of Theorem 6.4 is
as we wished to prove.
216
8. Tauberian Theorems
8. Summary
A typical Abelian theorem is the consistency or regularity one which states that a series or integral which converges is also summable, say in Abel‘s sense. A typical Tauberian theorem is a modified converse, a(t) dt, where say Littlewood’s, to the effect that the integral a(t) = O(l/t), t + 00, cannot be summable, in Abel’s sense, unless it converges. In proving this and similar results we used two approaches. In the first, a modification of Wiener’s method, we used integrals with very general kernels but imposed Tauberian conditions that were “ twosided ’”;in the second we used the Laplace kernel e-X’ only but imposed the less restrictive “ one-sided ” conditions. The results thus overlap in generality. It is noteworthy that no appeal to the theory of Fourier integrals was needed, nothing more recondite having been used than Ascoli’s selection theorem in the first method and Weierstrass’s approximation theorem for the second.
:5
EXERCISES 1. Show that Cesliro summability, as defined by (1.3) for series and
by (2.3) for integrals, is regular. 2. Illustrate Theorem 2.3 by the example a(t) = 1 - cos t , f ( x )
= 1/(x3
+ x).
+
3. Use the example a(t) = 1 t sin t to show that Theorem 2.4 would be false without hypothesis 2. 4. Prove the power series analog of Theorem 2.4 by use of that theorem :
1. l a k l < M , 2.
CakX-OD
k=O
n
someM, k = 0 , 1 , 2 ,....
A , 1- x
Cak-An,
k=O
x+l-. n4oo.
Exercises
5.
Use Theorem 2.1 to prove that
1
00
a(co) = A
0
e-xta(t) dt
-
A
-
X'
6. Prove that sin x # SO on (0, 00). [Hint: sin(x + n) - sin x = -2 sin x.]
7. Prove thatf(x)
E
SD on (0, 00) iff(x)
E B,
2 ?7
x+o+.
x f ( x ) E t.
+ b cos x E SD on (0, co). if a > 0 Show thatf(x) = x + cos x + cos(1og x ) E SD on (0,
8. Show that ax 9.
10. Show that f ( x ) in Exercise 9 is such that x f ( x ) [Hint: (xf(x))' = x(2
- sin x ) + cos x + a s i n
00).
1on (0, 03).
(: - log -
x
1.1
11. Usef(x) ofExercise 9 to determine whether Lemma 5.5 of Chapter 4 includes the present Theorem 6.1. 12. If h(t) = 1 on (1, e), h(t) = 0 elsewhere, show h(t) # U on (0, co). d Hint: Take a(t) = - sin(2 n log t), for example. dt
[
(L) som
1
13. By integrating formula 3, $2 of Chapter 1, prove rigorously that
tan-' 14. Prove that
1
- tan-'($
X
e-Xi
=
=
lo m
e-Xi dt
sin t
-dt, t
jo siny -d y , Y
x > 0.
x
> 0.
[Hint: Integrate (9.1) by parts.]
15. Show that Theorem 5.3 remains true if y 2 1 and hypothesis 2 is replaced by a(x) 2 - M on (0, a).
218
8. Tauberian Theorems
16. Show that if M is replaced by 0 in hypothesis 1 of Corollary 6.1 the conclusion follows trivially. dt
2 A(R)
17. Admitting equation (6.7) and Theorem 4.10 of Chapter 4, complete the proof of the prime number theorem by use of Exercise 7 and Theorem 6.1 of the present chapter.
Tauberian Theorems
1. Introduction
We recall that a series
is sermmable (A), in the sense of Abel, to the number A if the power series m
f(x)=
k= 0
akXk
converges for 1x1 < 1 and iff(1 -) = A . The classical theorem of Abel is a regularity or consistency result to the effect that the convergence of (1.1) to A implies its Abel summability to the same value. The example ak = ( - l)k shows that the converse statement is not true. It is this very fact that makes summability of divergent series a useful concept. A " corrected " converse is, however, true. For example, summability (A) implies convergence if one of the following conditions holds : 193
194
8. Tauberian Theorems
The first of these results is trivial as a consequence of the following inequalities : n
xakXksf(X),
O
k=O
n = O , 1 , 2 , ...,
Since the partial sums s, E t, they must approach a limit if f(1 -) is finite; hence summability implies convergence. The case B was first proved by A. Tauber [1897], case C by J. E. Littlewood [1910]. The latter result and closely related ones might be said to form the underlying theme of the present chapter. We recall further that series (1.1) is summable (C), in the sense of Cesaro, to the number A if lim
SO
n-t OCI
+ s1 + ..*+ s, n+l
= A.
This method of summability is also regular. (See Exercise 1 in this chapter.) The same example used above shows that summability (C) does not imply convergence. Again a “corrected” converse can be proved. For example, G . H. Hardy [1910, p. 3011 proved that condition C above is a suitable correction. We shall prove this later. Note that f(x)=
so + S I X
+ s2x2 + ...
1 S X +x2
+
...
so thatf(x) may be regarded as a weighted average, with weights xk, of all sk . Abel summability results in the limiting case when all weights are unity. Similarly, equation (1.3) asserts, in a sense, that the arithmetic average of “all” s k is A. These two examples are typical of
8.1. Introduction
195
summability theory. In general we say that a theorem is Abeliun if it asserts something about an average of a sequence from a hypothesis about its ordinary limit; it is Tuuberiun if conversely the implication goes from average to limit. Usually the latter results must have in the hypothesis some supplementary condition, like A, B, or C above. In this chapter we shall be dealing chiefly with Tauberian theorems involving integrals, which are the continuous analogs of series. After such results are proved it is then a fairly simple matter to obtain, as special cases, the classical ones about series. See $7. We conclude this section with the proof of a typical Abelian result, which will be useful later. Theorem 2.1.
Throughout this chapter we shall assume without further statement that u(t) E Lin (0, R) for every R > 0. Since
it is no restriction to assume that A = 0, u(t) = o(tY).Then to an arbitrary E > 0 there corresponds an R such that lu(t)l
Hence
< EtY,
R 5 t < co.
196
8. Tauberian Theorems
It follows that x -+ Of,
f ( x ) = o(x-y-'),
as we wished to prove. 2. Integral Analogs
In this chapter we shall be primarily concerned with Tauberian theorems involving Laplace integrals, and these may be regarded as continuous analogs of Tauber's original series theorem, or generalizations thereof. We say that the integral
is summable (A) to the value A if the Laplace integral f(x)=
(2.2)
I
m
0
e-%(t) dt
converges for x > 0 and iff(l-) = A . It is summable (C) if (2.3)
A(x) = Joxa(f) d t ,
lim
Our first result is the regularity theorem.
Theorem 2.1. 1. f ( x ) = 2. =>
JOm
m
0
e-%(t) dt,
0 < x < co;
a(t) dt = A
f(U+)
= A.
That is, the convergence of (2.1) implies its Abel summability to the same value. This result is an immediate consequence of Theorem 4.1 of Chapter 5. A converse of this, corresponding to Condition A of $1 follows.
8.2. Integral Analogs
197
Theorem 2.2. 1. f ( x ) =
m
J'0
0 < x < 00;
e-x'a(t) dt,
2. f ( O + ) = A ; 0 < t < co
3. a(t) 2 0 ,
a(t) dt
JOm
= A.
Analogous to the corresponding proof for series given in $1, this result follows easily from the inequality
A(R) = lim x+o+
loR
e "'a(t) dt -
5 l b f(x). x-o+
The example a(t) = cos t shows that hypothesis 3 cannot be omitted. For, in this case f ( x ) = x/(x' 1), f ( O + ) = 0, but the integral (2.1) diverges. Our next result is considered Abelian though it is a comparison of two kinds of averages. It states that Ceshro summability implies Abel summability.
+
Theorem 2.3. 1. f ( x ) = (2.4)
lorn0 < x < C x t a ( t )dt,
2. /:o(t)dt - A x ,
(2.5) If a(t) is an integral
then it is the integral
A
f(x)--,
X
x+
03
x+o+
00;
198
8. Tauberian Theorems
that is summable ( C ) by hypothesis 2 and which is summable (A) by (2.5). However, the theorem as stated, in terms of a(t) rather than b(t), suggests a different interpretation. The integral (2.1) diverges to + 00 ( A > 0), but the partial integrals A(x) become infinite in the prescribed way (2.4). The conclusion is that f(x) also becomes infinite, but at a resultant rate (2.5), as x + 0 . This result is essentially contained in Theorem 1. For, an integration by parts gives
+
f(x) = x
-
low e-XtA(t) dt,
x > 0,
and (2.4) becomes A(x) Ax, X + co. The conclusion of Theorem 1, y = 1, applied to the integralf(x)/x now yields the desired result. An example to show that the theorem is not reversible is a(t) = 1
(2.6)
+ sin t + t cos t , f(x)
-
1 x
=-
A(t) = t
+ t sin t
+ (x' 2x2 +
1)2'
Here f(x) l/x, x + 0 + , but A (t)/t approaches no limit as t + co. Accordingly, to obtain a corrected converse we must expect to impose some additional condition on a(t). It turns out that boundedness is a suitable one. Note that a(t) as defined by (2.6) is not bounded.
Theorem 2.4. W
1. f(x> =
J0
e-xta(t)
dt,
2. a(t) E B (bounded), 3. f(x)--, (2.7)
=r
1:
A X
a(t) dt
o < x < co; 0 < t < 00;
x+o+
-
Ax,
x + 00.
We shall obtain this result in the next section as a special case of a much more general theorem. But let us illustrate it here by the example
8.3. A Basic Theorem
199
The Laplace transform of a(t) is (Exercise 14) (2.8)
1
1
f(x) = - tan-’ X x
7
I
-
x+o+.
2x’
Since a(t) E B we may apply the theorem to obtain (2.7) with A But this may be verified directly since Joxa(f)dt = r ( x 0
sin y
.x
sin y
Y
O
Y
- y ) -d y = x J -dy
x-rm X
a(t) dt =
I
c0
O
sin y
7I
Y
2
= 42.
+ cos x - 1,
-d y = -.
Exercise 3 shows that hypothesis 2 is essential here. 3. A Basic Tauberian Theorem
We observe that hypothesis 3 of Theorem 2.4 may be written as
if g(t) = e - t and that the conclusion (2.7) takes the same form if g(t) is replaced by h(t) = 1 on (0, l), h(t) = 0 on (1, 00). Thus Theorem 2.4 states that, under the restriction of boundedness on a(& if (3.1) holds for one “ kernel ’’ g(x) it also holds for at least one other kernel h(t). Our basic theorem will show that the conclusion will hold for many kernels h(t) even if g(t) is not e-t. The essential property of e - * for our purpose is contained in the familiar uniqueness property of the Laplace transform: the vanishing of the generating function implies that of the determining function. We dignify this property in a kernel g(t) by a notation as follows. Definition 3.1. A function g(x) E U (has the uniqueness property) on (0, co)if g(x) E L on (0, co) and if the equation Jomg(;)a(t)
dt
E
0,
0 < x < co
for a(t) E B. C implies a(t) = 0, 0 < t < co.
8. Tauberian Theorems
200
Clearly an exponential change of variables converts the integral (3.1) into a convolution, the range of integration becoming the whole real axis. Our uniqueness property takes the following form for the interval (-a, a). (-
Definition 3.2. A function g(x) E U on (- co,00) if g(x) E L on co) and if the equation
rm
00,
g(x - t) a(t) dt
= 0,
0
for a(t) E B. C implies a(t) = 0, - co < t < co. Note that if g(x) E U on (0, 03) then g(e")e" E U on (- 00, m). For example e-" E U on (0, 00) and exp(-e")e" E U on (- co, 00). On the other hand a function g(x) which is constant over a finite interval and is zero elsewhere 4 U on (- 00, 00). This can be seen'by choosing a(t) as a suitable periodic function. (See Exercise 12 of this chapter.) We now state our basic Tauberian Theorem as follows.
Theorem 3.1. 1. g(x)E
u,
2. a(x) E B ,
3. h(x) EL, 4.
Jm -m
(-a, a); (-a, co);
(-a, m);
- t)a(t)dt
3
11m
g(x - t)a(t) d t --t A +A
m
m
m
g(t) dt,
h(t) d t ,
x+m
x + m.
By considering the function a(t) - A we see that there is no loss of generality in assuming A = 0. Set G=g*a,
H=h*a.
From the assumption G(m) = 0 we wish to prove H(m) = 0. Assuming the contrary, there must exist 6 > 0 and a sequence x,,tending to co with n such that H(xJ I > 6. Now observe that G * h = H * g = g * h *a. (3.2)
I
+
This follows by Fubini's theorem in the presence of hypotheses 1, 2, 3.
8.3. A Basic Theorem
Now consider the sequence = H(x
s,(x)
201
+ x,).
We shall show that it is bounded and equicontinuous on - co < x < co. For, if M is an upper bound for I a(t) 1, then
-co < x < co,
IH(x)I 5 M I m Ih(t)l dt, -m
I s,(x) - s,(y) I 5 M J”- m 1 h(x + X, - t ) - h(y + X, - t ) [ m
dt
=~(l),
y-x40.
The latter result follows from a classical result of Lebesgue theory ; see N. Wiener [1933, p. 141. By Ascoli’s lemma (see below) we now select from the functions sn(x) a subset s,(x) which tends to a limit s(x), continuous on (- co, co). By Lebesgue’s limit theorem, applicable since H E B, g EL, we have lirn
k-ao
(3.3)
1- s,,(x
= lirn k-m
m
m
- t ) g ( t ) dt
= lim k+m
m
m
J”-ao G(x + xnk- t)h(t) dt =
H(x
1
m
W
-m
+ x,,
- t)g(t) dt
s(x - t ) g ( t ) dt = g
* s.
Here we have used equation (3.2). Since G(m) = 0, another application of the Lebesgue limit theorem shows that the limit (3.3) is zero. Since s(x) E B.C we infer from hypothesis 1 and Definition 3.2 that s(x) -= 0. But Is(0)l = lirn Is,,(O) I = lim IH(x,,,)l 2 6 > 0. k+m
k-m
The contradiction shows that H(co) = 0, as we wished to prove. For the reader’s convenience we prove the form of Ascoli’s lemma which we have used.
Lemma 3.1. 1. {s,(x)}~ is equicontinuous on - co < x < co;
2.
Is,(x)l < M ,
+ A.
lim s,(x) k+ m
B. s ( x ) E C ,
n
=
= s(x),
I , 2, ..., -co < x <
00,
some M
some integers nk , some s(x);
-w
202
8. Tauberian Theorems
Arrange all real rational numbers in a sequence xl, x 2 ,x3, . . . . From the bounded double sequence s,(x,,) we may pick, by the familiar diagonal process, a subsequence snk(xm)such that (3.4)
lim snk(x,)= s(x,),
k- m
m = 1,2, 3,
. . ..
See, for example, D. V. Widder [1946, p. 261. The function s(x) is defined by the limit (3.4) at all rational points. Now let y be irrational and let E > 0 be arbitrary. By hypothesis 1, if x , is sufficiently near to y ,
I snk(y) - S n k ( x m ) I < E , IS,,(Y)
< E,
- t9,(x,)l
1,2937 . . ., j = 1,2,3, ....
k
=
Such an x , exists since y is a limit point of rational points. By Cauchy's limit criterion equation (3.4) implies the existence of a number K such that k >K, j > KISnk(Xrn) - snl(Xrn)l < E , By the triangle inequality
I S,,(Y)
- s,,(Y>
I <3 ~ ,
k >K, j >K,
from which we may infer the existence of
(3.5)
lim
k+m
S",(Y> = 4 Y > .
It remains to show only that s(x), now defined for all x, is continuous. By the equicontinuity of the sequence snk(x)we have for a given E the existence of a number 6, independent of nk, such that
1
snk(x)
- snk(y)
Is(x) - 4 Y ) I
I < 8,
s
E,
1 x - Y I < 6, Ix - Yl < 8.
Thus the lemma is proved. We point out that Ascoli's result is usually stated for compact regions, in which case the limit (3.5) is uniform. That the approach need not be uniform on (- 00, co)may be seen by considering sn(x) = sin (x/n). Here s(x) = 0, but any inequality of the form Is,(x)I < E < 1 must fail (at x = ns1/2, for example). The previous theorem, restated for the interval (0, co) becomes the following.
8.4. Hardy and Littlewood Theorems
203
Theorem 3.2.
u,
(0, co);
2. u(x) E B,
(0, co);
3. h(x) E L ,
(0, co);
1. g ( x ) E
We may specialize this result by choosing g(x) = e - x x a - l and h(x) = xP-' on (0, l), h(x) = 0 on (1, co). The hypotheses 1 and 3 are evidently satisfied if a > 0, > 0.
Theorem 3.3
2. a(x) E B, 3. f ( x )
(0, a);
- ma) XU
'
x -+
o+
In particular if a = /? = 1, this result reduces to Theorem 2.4, the proof of which we had deferred. 4. Hardy's and Littlewood's Integral Tauberian Theorems
As we noted in $1, Hardy proved first that if ak = O(l/k), k 3 co, then the series (1.1) cannot be summable in the sense of CesAro unless it converges; Littlewood (1912, p. 434) proved the same result for Abel summability. In this section we shall prove the integral analogs of these results, deriving therefrom i n 97 the classical series result. We need first a new definition.
8. Tauberian Theorems
204
Definition 4. A function f ( x ) E SO (slowly oscillating) on (0, co) if lim M
Y ) -f(x)l
=0
when y and x + 00 in such a way that y / x + 1. For example, if f ( x ) f ( x ) E SO there. For,
I f ( Y ) - f ( x >I 5
and x l f ’ ( x ) l < M on (0, co) then
E C’
lpv) I s
log($
= o(l),
L 1 .
X
Thus sin (log x ) E SO on (0, 00). On the other hand, sin x $ SO on (0, co). (See Exercise 6 of this chapter.)
Theorem 4.1.
1. a(x) E so, -x
2. Jo a(t) dt
*
-
(0, co);
AX,
co
X +
a(co) = A .
As usual, we may assume A = 0. From hypothesis 1 we see that to an arbitrary E > 0 there correspond numbers K and S such that la(y) - a(x)I
y > K, x > K,
< E,
That is, for x - 6x < y < x
+ 6x
a(x) - E
Integration gives
[a(x) - ~ 1 2 < 6 ~
Now let x Thus
+ 00.
x(li6)
x(1-a)
lim
X+ 00
a(y) dy
< [a(x) + E ] 2Sx.
The integral (4.1) is o(x), x [a(x)
- E l 2s 2 0,
-&
- 11 < 6.
< a ( y ) < a(x) + E .
1
(4.1)
:1
l&ij
[a($
-,co, by
+ E l 2s 2 0,
5 lim a(x) 5 ihi a(x) 5 E .
The conclusion is now immediate.
hypothesis 2.
8.4. Hardy and Littlewood Theorems
205
Hardy's integral Tauberian theorem is an immediate consequence of this result.
Theorem 4.2. (0, GO);
1. xa(x) E B,
2. A ( x ) =
3. (4.2)
Jx 0
lx dt ; 0
a(?)
A ( t ) d ?- A x ,
J,
~ ( tdt)
x-+GO
= A.
Hypothesis 1 corresponds to condition C of $1 and hypothesis 3 states that the integral (4.2) is summable (C) to the value A . We apply Theorem 4.1 to the function A(x). It belongs to SO since
Here M is a bound for 1 xa(x) 1. The conclusion of Theorem 4.1 is A(m) = A , equivalent to (4.2). The proof of Littlewood's integral Tauberian theorem is slightly more complicated. Theorem 4.3.
(4.3) +-
m
jo a(t) = A .
By use of Theorem 2.4 we show first that, in the presence of hypothesis 1, the Abel summability of the integral (4.3) implies its Ceshro
206
8. Tauberian Theorems
summability, and we can then apply Theorem 4.2. An obvious integration by parts enables us to write hypothesis 2 as
lom e-"'A(t) dt - A x ,
x +0
+.
To prove that A ( x ) E B, consider the difference a(t)[1 - e-'IX]
dt -
j
m
e-'lxu(t)
dt.
X
By hypothesis I
5M
i;"
+M
dt
2M.
Since f ( x ) E B on (0, co) by hypothesis 2, the same must be true of A(x). Hence if we apply Theorem 2.4 to A(x) we find that
j;A(t) dt - A x ,
x -+ co.
That is, the integral (4.3) is summable ( C ) to A , and the convergence of (4.3) now follows from Hardy's theorem. As an illustration take a(t) = (sin t)/t. Its Laplace transform is tan-'(llx) which tends to 742 as x + O + . Obviously ta(t) E B, so that
somSF dt
= 71 -
2'
5. One-sided Tauberian Conditions
In Hardy's and Littlewood's theorems the condition u(x)x EB can be replaced by a one-sided condition in which a(x)x is bounded below (or above) only. With a view to proving this we first establish several results of J. Karamata [1931, p. 271.
8.5. One-sided Conditions
Theorem 5.1. 1. g ( x ) = O , O S x < c < l ;
2. E > 0 ,
* (5.1) (5.2)
g(x)EC,
207
c5xg1;
y>o
There exist polynomials p ( x ) , P ( x ) such that p(x) < g(x) < P(x), Jol
[log(;)]y-
05x5 17
b ( x ) - p ( x ) ] d x < E.
For an arbitrary positive number q we can determine h(x) E C on 0 5 x 5 1 such that g(x)
5 h(x),
0 5 x 2 1.
For example, if g ( c + ) > 0 (the only case needed later) we may take h(x) = g(x) except in a small interval (c - 6, c) where h(x) = g(c+)(x - c + 6)/6. The value of the integral (5.3) is less than
Hence inequality (5.3) may be satisfied by choice of 6. By the Weierstrass approximation theorem we determine a polynomial Q(x) such that Ih(x) - Q(x)I < q on 0 S x 5 1, and we define P(x) = Q(x) q. Then g(x) 5 h(x) < P(x) and
+
x [(P - Q) +
I Q - h I + ( h - g)l dx <
+ Y ~ ( Y )+ V .
Similarly, there is a polynomial p(x) such that p(x) < g(x) on (0, 1) and such that
208
8. Tauberian Theorems
+
Adding inequalities (5.4) and (5.5) and choosing g = &/(4r(y) 2), we have (5.2). Hypothesis 1 could be very considerably relaxed if desired; see D. V. Widder [1941, p. 1891. Theorem 5.2.
0 < x < co;
2. a(x) 2 0 ,
3. g(x)=O, 4. f ( x )
osx
A
y>o,
xy '
N -
g(x)EC, c l x S 1 ;
x+o+.
x+o+.
We begin by determining the polynomials p and P of Theorem 5.1, corresponding to a given E . Since a(x) 2 0 we have (5.7)
/
e-"'p(e-"')u(t) dt
m
0
e-"'g(e-"')a(t) dt
5
1
W
0
e-"'P(e-"')a(t) dt,
If we replace x by (n + 1)x in hypothesis 4 we obtain
Thus the conclusion (5.6) is valid if g(x) is replaced by X" and hence by any polynomial. From (5.7) and (5.9) we have e-'p(e-')tY-' dt
5 l b xT(y) x-o+
S fi;;; x T ( y ) / x+o+
m
e-X' g(e-"')a(t) dt 0
00
0
e-"'g(e-"')a(t)
dt
5A
/
m
0
e-'P(e-')ty-'
dt.
8.6. One-sided Littlewood Theorem
209
By (5.2) the two extremes of this inequality differ by AE.Hence the two terms in the middle must be equal to each other and to the middle term of (5.8). That is, (5.6) is established. We now specialize the function g(x) to obtain the following Tauberian theorem. Theorem 5.3.
3. f ( x )
-
A
-
xy
'
y>o,
x-bo+
Except for the present one-sided Tauberian condition 2 this result would be included in Theorem 3.3. To prove it, choose g(x) = l / x on (l/e, l ) , g(x) = 0 on (0, l/e) in Theorem 5.2. Then (5.6) becomes
which is equivalent to (5.10). 6. One-sided Version of Littlewood's Integral Theorem
To obtain a one-sided version of Theorem 4.3 we need several preliminary results, of interest in themselves. We introduce a definition. Definition 6. A function f ( x ) E SD (slowly decreasing) on (0,co) if (6.1)
lim M Y ) - f W l 2 0,
when y and x become infinite in such a way that y > x and y/x + 1 .
8. Tauberian Theorems
210
For example, any increasing function belongs to this class. Or,
f(x) E SD if xf’(x) > - M for some constant M . For,
(;),
f(Y) - A x ) > -Mlog
Y > x,
from which (6.1) is immediate. We now prove a one-sided version of Theorem 4.1.
Theorem 6.1. 1. a(x) E SD,
2.
/ox
a(t) dt
-
Ax,
x + co
a(w) = A.
=>
Assume A
(0, 00);
= 0.
For an arbitrary
> 0 we have from (6.1) that x < y < (1 3. 6)x. E
a(x) - E < a(y), (6.2) Integrating with respect to y and using hypothesis 2, we obtain
lim a(x)
7
(6.3)
x-t m
5 E.
Rewrite (6.2) as
and integrate with respect to x :
This with (6.3) shows that a(co) = 0, as we wished to prove. Corollary 6.1. 1.
xa(x) > - M ,
2.
JOm
a(x) dx
someM, O < x < co;
is summable (C)to A
6.8. One-sided Littlewood Theorem
211
This is a one-sided version of Theorem 4.2. It follows at once by observing that A@) now satisfies the conditions of Theorem 6.1. We prove next a result of E. Landau [1929,p. 581.
Theorem 6.2.
1.
osx
f(X)EC2,
2. f ( 0 ) = A ; -M 3. f”(x) > x2
*
’
someM, O < x < c o
(3
f’(x)=o - ,
x-+o+.
From Taylor’s theorem we have for 0 < 6 c I
Now let x obtain
-+
0
+ , noting that f ( x + ax) - f ( x )
-+
0 by hypothesis 2.We
-M6 M6 < lim x f ‘ ( x ) 5 G xf’(x) 5 -. 2(1 - S ) 2 =xz+ x + o + 2 Since 6 is arbitrary we obtain at once the desired conclusion. We derive next a consequence of Theorem 4.3, using a modification of the first hypothesis thereof.
Theorem 6.3. 1. B(x) =
jxta(t) dt = o(x), 0
2. f ( x ) = Jorn e-x‘a(t) dt -+ A , (6.4)
=>
jo* a(t) dt = A .
x
-+
x
co ;
-+
o+
212
8. Tauberian Theorems
This states that hypothesis 1 is an alternative condition to Littiewood's for the Abel summability of the integral (6.4) to imply its convergence. Assume A = 0. Integration by parts gives (6.5) Jme-xf 1
3 dt = B(l)e-" - x Jme-"' 9 dt + 1e -%(t) t2 t m
1
1
dt.
The determining function B(t)/t2of this first integral is U ( l / t ) by hypothesis 1. Hence we may apply Theorem 4.3. Its limit as x 0 + , from the right-hand side of (6.5) is --f
B(l) -
For, lim x
x+o+
1
1
0
a(t) dt.
f,*e - X fB(t) dt = 0
by Theorem 2.1, and
lim fpe-.i2ct, dt
=
- lim
Jol
e-*'a(t) dt
1
=
- l 0 u ( t )dt
by hypothesis 2. Thus the conclusion of Theorem 4.3 yields
jlm $?dt = B ( l ) - j a(t) dt, 1
0
=B(l)
(6.7)
+
m 1
a(t) dt.
To obtain (6.7) we have integrated integral (6.6) by parts. Equation (6.4), with A = 0, is now immediate by comparing (6.6) with (6.7). We can now obtain the desired one-sided version of Littlewood's theorem very easily.
Theorem 6.4.
2. f ( x ) =
loebxta(t)dt JOm
a(t) dt
+A ,
= A.
x + O+
8.7. Classical Series Results
2 13
From hypothesis 1 we have
By Theorem 6.2f'(x) = o(l/x), x +O+. Thus Jbme-"'[M
+ ta(t)] dt = M-f ' ( x ) X
-
M
-, X
x -+
O+
Since M + ta(t) > 0 we may apply Theorem 5.3 to this integral to obtain
Jb'
[ M + tu(t)] dt
N
Mx,
x + 00.
This implies that hypothesis 1 of Theorem 6.3 is valid. The conclusion of that theorem is the desired result here. 7. Classical Series Results
We have hitherto proved the integral analogs of the Tauberian theorems about series mention in $1. It is now possible to obtain these discrete results, not by analogy, but as direct consequence of the continuous cases. We do so in a few instances, using the more general onesided versions. First, the discrete analog of Theorem 2.4 or of Theorem 5.3 (y = 1): Theorem 7.1. (7.1)
1.
2.
m
xake-kx-tA,
k=O
n
-M
x+O+;
someM, n = 0 , 1 , 2 ,...
=>
That is, if the sequence of partial sums of a series is bounded on one side it cannot be summable in the sense of Abel unless it is also summable in the sense of Ceshro. It is obvious that for y = 1 in Theorem 5.3 the lower bound zero of hypothesis 2 may be replaced by any number -M. (See Exercise 15 of this chapter.)
214
8. Tauberian Theorems
Define a function A(x) so that A(0) = 0 and A(x) =
c a, > - M ,
[XI
k=O
x > 0.
If the sum of the series (7.1) isf(x), we have
1 e-”‘A(t) dt, 03
W
f ( x ) = Jo e-”‘dA(t) = x
0
x
> 0.
Thus by hypothesis 1 j o r n e - X t A ( t ) d t -A- ,
x-bo+.
X
The conclusion of Theorem 5.3, with x replaced by n
+ 1, is
This is the desired result. We turn next to the series form of Hardy’s theorem.
Theorem 7.2.
n
n
Observe k s t that hypothesis 2 implies that sn= o(n) and hence that an = o(n), n -+ 00. Define A(x) as in the previous theorem and apply Theorem 6.1 to it. Set m = [ y ] and n = [x], y > x. Then
8.7. Classical Series Results
215
From these inequalities it is clear that A(x) E SD on (0, co). Further, if n 1 S x < n + 2
+
Then by hypothesis 2
j; A ( t ) dt
N
Ax,
x
+
co.
The conclusion of Theorem 6.1 is A(0)=
m
Uk k= 0
= A,
as we wished to prove. The series version of Littlewood’s theorem follows. Theorem 7.3. 1.
ak
-M
someM, k = 0 , 1,2, ...;
>k + 1’
W
We apply Theorem 6.4 defining a(x) by U(x)=Uk,
ksx
k = 0 , 1 , 2 ,....
From hypothesis 1 it is clear that a(x) > - M / x for 1 4 x < co. The conclusion of Theorem 6.4 is
as we wished to prove.
216
8. Tauberian Theorems
8. Summary
A typical Abelian theorem is the consistency or regularity one which states that a series or integral which converges is also summable, say in Abel‘s sense. A typical Tauberian theorem is a modified converse, a(t) dt, where say Littlewood’s, to the effect that the integral a(t) = O(l/t), t + 00, cannot be summable, in Abel’s sense, unless it converges. In proving this and similar results we used two approaches. In the first, a modification of Wiener’s method, we used integrals with very general kernels but imposed Tauberian conditions that were “ twosided ’”;in the second we used the Laplace kernel e-X’ only but imposed the less restrictive “ one-sided ” conditions. The results thus overlap in generality. It is noteworthy that no appeal to the theory of Fourier integrals was needed, nothing more recondite having been used than Ascoli’s selection theorem in the first method and Weierstrass’s approximation theorem for the second.
:5
EXERCISES 1. Show that Cesliro summability, as defined by (1.3) for series and
by (2.3) for integrals, is regular. 2. Illustrate Theorem 2.3 by the example a(t) = 1 - cos t , f ( x )
= 1/(x3
+ x).
+
3. Use the example a(t) = 1 t sin t to show that Theorem 2.4 would be false without hypothesis 2. 4. Prove the power series analog of Theorem 2.4 by use of that theorem :
1. l a k l < M , 2.
CakX-OD
k=O
n
someM, k = 0 , 1 , 2 ,....
A , 1- x
Cak-An,
k=O
x+l-. n4oo.
Exercises
5.
Use Theorem 2.1 to prove that
1
00
a(co) = A
0
e-xta(t) dt
-
A
-
X'
6. Prove that sin x # SO on (0, 00). [Hint: sin(x + n) - sin x = -2 sin x.]
7. Prove thatf(x)
E
SD on (0, 00) iff(x)
E B,
2 ?7
x+o+.
x f ( x ) E t.
+ b cos x E SD on (0, co). if a > 0 Show thatf(x) = x + cos x + cos(1og x ) E SD on (0,
8. Show that ax 9.
10. Show that f ( x ) in Exercise 9 is such that x f ( x ) [Hint: (xf(x))' = x(2
- sin x ) + cos x + a s i n
00).
1on (0, 03).
(: - log -
x
1.1
11. Usef(x) ofExercise 9 to determine whether Lemma 5.5 of Chapter 4 includes the present Theorem 6.1. 12. If h(t) = 1 on (1, e), h(t) = 0 elsewhere, show h(t) # U on (0, co). d Hint: Take a(t) = - sin(2 n log t), for example. dt
[
(L) som
1
13. By integrating formula 3, $2 of Chapter 1, prove rigorously that
tan-' 14. Prove that
1
- tan-'($
X
e-Xi
=
=
lo m
e-Xi dt
sin t
-dt, t
jo siny -d y , Y
x > 0.
x
> 0.
[Hint: Integrate (9.1) by parts.]
15. Show that Theorem 5.3 remains true if y 2 1 and hypothesis 2 is replaced by a(x) 2 - M on (0, a).
218
8. Tauberian Theorems
16. Show that if M is replaced by 0 in hypothesis 1 of Corollary 6.1 the conclusion follows trivially. dt
2 A(R)
17. Admitting equation (6.7) and Theorem 4.10 of Chapter 4, complete the proof of the prime number theorem by use of Exercise 7 and Theorem 6.1 of the present chapter.