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26 Convolution of Functions
26 Convolution of Functions Let f , g be two continuous complex-valued functions in R”,having compact support. The convolution of f and g is the function, also defined in R”, (26.1)
(f* g)(x) = J-R”f(x - Y)g(Y) 4J= J-R“f(Y)g(x- Y ) dr.
It is clear, on this definition, that there is no need for both f and g to have compact support: the integrals defining f * g have a meaning if either one of the “factors” f or g has compact support while the other has an arbitrary support. Under this assumption, it is immediately seen that the function f * g is continuous. Furthermore, if both supp f and supp g are compact, so is supp( f * g ) . If now we release the requirement that f and/or g have compact support, it is clear that we must impose a condition ensuring that, for every x E R”,the function of y , f ( y ) g ( x - y ) , be integrable. This demands that the two functions f and g be locally integrable and that the growth at infinity of one of them be “matched” by the decay of the other. What we have said earlier is an illustration of this situation: if the growth at infinity off, say, is arbitrary, then the decay of g must also be arbitrarily fast, which can only mean that g vanishes outside a compact set. But other conditions, less “lax” on f and less restrictive on g, may easily be imagined. Suppose for instance that f grows slowly at infinity, i.e., slower than some polynomial, and that g decreases rapidly at infinity, i.e., faster than any power of 1/1 x I. Then obviously f * g is well defined by Eq. (26.1). As will now be shown, one may strengthen the condition on f and weaken the one on g in such a way that the two conditions come to coincide, and in fact are also shared by the resulting function f *g; and the conditions are nothing else but that f and g be integrable!
THEOREM 26.1. Let p , q, and Y
< +m
(26.2)
Y
be three numbers such that 1
and such that
llr
= U/P)
+ (l/!?) - 1-
278
< p , q,
279
CONVOLUTION OF FUNCTIONS
Then,for all pairs of continuousfunctions with compact support in Rn, f,g ,
we have
Proof. Set h(x) = ( f * g ) ( x ) = JRn f ( x - y)g(y) dy. Set s = p(1 - l / q ) . By Holder's inequalities (Theorem 20.3), we obtain
I Wl <
1s
If(.
I/@
-y)l(l-s)a I g o l q d y l
1I lflSIlL~*>
where q' is the conjugate of q, i.e., q'-l = 1 - q-l. Observing that sq' = p , we obtain (26.4)
I h(x)lq < llf113
J If(.
-Y)l(l--s)q I g(y)l*dY.
-
At this point, we make use of the following general fact: let t f ( t ) be a continuous function from Rn into some Banach space E, which has compact support. We may then define its integral JRn f ( t )dt, say by considering Riemann sums; the value of this integral is an element e of E. We have (26.5)
We have denoted by 1) 1) the norm in E. Our statement follows immediately from the "triangular" inequality for the norm applied to the finite Riemann sums approaching the integral (also observe that t 1) f(t)ll is a nonnegative continuous function with compact support in Rn).We apply this to the function
-
Y
- If(.
- Y)l(l-s)O I g(Y)lq,
regarded as a function of y E Rn,obviously continuous with compact support, into some space La (with respect to the variable x). By applying (26.5) together with (26.4), observing that h(x) is a continuous function with compact support in Rn,we obtain
But this can be rewritten as
280 We choose obtain
[Part I1
DUALITY. SPACES OF DISTRIBUTIONS (Y
= r/q
and take the qth root of both sides of (26.6). We
I1 h llL'
< Ilfllb
I1g llLq *
l l m 1 ~ * ) r
This is nothing else but (26.3) if we observe that (l-s)r=
(
P
1
l-p++=pt
(5
4
1
+--1)=p, 4
by (26.2). Q.E.D. By using the density of the space e ( R n )of continuous functions with compact support in the spaces La(Rn),01 < +a,one may then prove the following consequence of Theorem 26.1 (when I < +a;when r = +m, one applies Holder's inequalities):
COROLLARY 1. Iff E Lp, g EL*, then
defines an element of Lr(p,q, r as in Theorem 26. I), denoted by f * g ; we have (26.7)
Ilf*gll,r<
IlfllL~IlgllL~.
From this one, the following results are easily derived:
COROLLARY 2. Let p be a number such that 1 f E L', the mapping
< p < +w. For
every
g -f * g
is a continuous linear map of Lp into itself, with norm
f llLl .
COROLLARY 3. The convolution
is a bilinear mapping of L1 x L' into L1;we have: Ilf*g
llL'
< Ilf
llL1
llg llL1 .
One often summarizes the content of Corollary 3 by saying that L1 is a conwolution algebra. Let A be an arbitrary subset of Rn, and f and g two continuous functions in Rn, one of which has a compact support. Let fA be the
Chap. 26-41
28 1
CONVOLUTION OF FUNCTIONS
function equal to f in A - s u p p g and to zero everywhere else. Then we have obviously, for all x E A,
We may summarize this as follows:
26.1. Let f , g be continuous functions in Rn,one of which has a compact support. Let A be a subset of R". The values of the convolution f * g in the set A do not depend on the values o f f in the complement of the set PROPOSITION
A
-
suppg
= {x E
Rn;x
= x'
- x"
for some x'
E
A', x" E suppg}.
We have already taken advantage of the fact that, if both f and g have compact support, so does f *g. This may be regarded as a corollary of the following result:
PROPOSITION 26.2. Let f , g be as in Proposition 26.1. W e have (26.9)
supp(f* g) C suppf
+ suppg
(vector sum).
+
Proof. Let x belong to the complement of supp f supp g ; then, for y E supp g, x - y belongs to the complement of supp f , hence
COROLLARY. If both supp f and supp g are compact, supp( f compact.
* g) is also
Indeed, if K and H are compact subsets of a HausdorfT TVS E (here R"), K H is also a compact subset of E, as it is the image of the product K x H, compact subset of E x E, under the continuous mapping (x, y ) x y. We proceed now to study convolution from the viewpoint of differentiability. Suppose that f and g are two V1functions, one of which has compact support in Rn.Then it follows immediately from Leibniz' rule .for differentiation under the integral sign that
+
- +
282
DUALITY. SPACES OF DISTRIBUTIONS
[Part I1
is a differentiable function (at every point x) and that
( W A f* g ) = ( ( V x J f )
*o"
=f
* ((a/%)g),
i = 1, ...,n.
In fact, combining this with the fact that the convolution of two continuous functions, one of which has compact support, is a continuous function, we see that f * g is a W function in R". Furthermore, if we apply (26.8) with af/axi instead off, we derive the fact that, for all x E A,
where B = A - supp g. Of course, the right-hand side of the above inequality may very well be infinite. But it will be finite whenever supp g and A are both compact sets. This implies easily the following result, which is a particular case of a more general fact: PROPOSITION 26.3. Let m be an integer, 0 compact support, the convolution
f
< m < +a.If g EL^
has
-f*g
is a continuous linear map of Vm(Rn)into itself.
COROLLARY. Let m, g be as in Proposition 26.3. The convolution f is a continuous linear map of VF(Rn)into itself.
- *g f
Proof of Corollary. In view of the general properties of LF-spaces, it suffices to prove that, for every compact subset K of Rn, the restriction of the mapping f f * g to V T ( K )is continuous, as a map of %?F(K) into VT(Rn)(see Proposition 13.1). But in view of Proposition 26.2, it maps V T ( K ) into VT(K supp g), and on both these spaces the topology induced by Vr(R") is the same as the one induced by P ( R n ) . The corollary follows then directly from Proposition 26.3.
-
+
Exercises 26.1. Let p be a Radon measure, and f a continuous function with compact support in R". Prove that (P
is a continuous function in R".
*f)(x)
=
s
f(x - Y ) M Y )
Chap. 26-61
CONVOLUTION OF FUNCTIONS
283
26.2. Let f,g be two locally L' functions in the real line which vanish identically in the negative open half-line { t E R'; t < O}. Prove that
(f*g)(x)
=
J'f(x
- t ) g ( t ) dt
is a locally L1function in R' vanishing for x < 0. 26.3. Compute the Fourier transform off * g when f,g E Y(R"). 26.4. Let j,g be two continuous functions in R",one of which has compact support. Prove Leibniz' formula:
<
where p , q E N" and q p means q j < pi for all j = 1 ,..., n. 26.5. Let .f be a continuous function with compact support in R". Prove that, if g is exp(), 5 E C";resp. a function which can a polynomial (resp. an exponential x be continued to the complex space C" as an entire analytic function), the same is true of the convolution f * g.
-
(f* g)(x) = J-R”f(x - Y)g(Y) 4J= J-R“f(Y)g(x- Y ) dr.
It is clear, on this definition, that there is no need for both f and g to have compact support: the integrals defining f * g have a meaning if either one of the “factors” f or g has compact support while the other has an arbitrary support. Under this assumption, it is immediately seen that the function f * g is continuous. Furthermore, if both supp f and supp g are compact, so is supp( f * g ) . If now we release the requirement that f and/or g have compact support, it is clear that we must impose a condition ensuring that, for every x E R”,the function of y , f ( y ) g ( x - y ) , be integrable. This demands that the two functions f and g be locally integrable and that the growth at infinity of one of them be “matched” by the decay of the other. What we have said earlier is an illustration of this situation: if the growth at infinity off, say, is arbitrary, then the decay of g must also be arbitrarily fast, which can only mean that g vanishes outside a compact set. But other conditions, less “lax” on f and less restrictive on g, may easily be imagined. Suppose for instance that f grows slowly at infinity, i.e., slower than some polynomial, and that g decreases rapidly at infinity, i.e., faster than any power of 1/1 x I. Then obviously f * g is well defined by Eq. (26.1). As will now be shown, one may strengthen the condition on f and weaken the one on g in such a way that the two conditions come to coincide, and in fact are also shared by the resulting function f *g; and the conditions are nothing else but that f and g be integrable!
THEOREM 26.1. Let p , q, and Y
< +m
(26.2)
Y
be three numbers such that 1
and such that
llr
= U/P)
+ (l/!?) - 1-
278
< p , q,
279
CONVOLUTION OF FUNCTIONS
Then,for all pairs of continuousfunctions with compact support in Rn, f,g ,
we have
Proof. Set h(x) = ( f * g ) ( x ) = JRn f ( x - y)g(y) dy. Set s = p(1 - l / q ) . By Holder's inequalities (Theorem 20.3), we obtain
I Wl <
1s
If(.
I/@
-y)l(l-s)a I g o l q d y l
1I lflSIlL~*>
where q' is the conjugate of q, i.e., q'-l = 1 - q-l. Observing that sq' = p , we obtain (26.4)
I h(x)lq < llf113
J If(.
-Y)l(l--s)q I g(y)l*dY.
-
At this point, we make use of the following general fact: let t f ( t ) be a continuous function from Rn into some Banach space E, which has compact support. We may then define its integral JRn f ( t )dt, say by considering Riemann sums; the value of this integral is an element e of E. We have (26.5)
We have denoted by 1) 1) the norm in E. Our statement follows immediately from the "triangular" inequality for the norm applied to the finite Riemann sums approaching the integral (also observe that t 1) f(t)ll is a nonnegative continuous function with compact support in Rn).We apply this to the function
-
Y
- If(.
- Y)l(l-s)O I g(Y)lq,
regarded as a function of y E Rn,obviously continuous with compact support, into some space La (with respect to the variable x). By applying (26.5) together with (26.4), observing that h(x) is a continuous function with compact support in Rn,we obtain
But this can be rewritten as
280 We choose obtain
[Part I1
DUALITY. SPACES OF DISTRIBUTIONS (Y
= r/q
and take the qth root of both sides of (26.6). We
I1 h llL'
< Ilfllb
I1g llLq *
l l m 1 ~ * ) r
This is nothing else but (26.3) if we observe that (l-s)r=
(
P
1
l-p++=pt
(5
4
1
+--1)=p, 4
by (26.2). Q.E.D. By using the density of the space e ( R n )of continuous functions with compact support in the spaces La(Rn),01 < +a,one may then prove the following consequence of Theorem 26.1 (when I < +a;when r = +m, one applies Holder's inequalities):
COROLLARY 1. Iff E Lp, g EL*, then
defines an element of Lr(p,q, r as in Theorem 26. I), denoted by f * g ; we have (26.7)
Ilf*gll,r<
IlfllL~IlgllL~.
From this one, the following results are easily derived:
COROLLARY 2. Let p be a number such that 1 f E L', the mapping
< p < +w. For
every
g -f * g
is a continuous linear map of Lp into itself, with norm
f llLl .
COROLLARY 3. The convolution
is a bilinear mapping of L1 x L' into L1;we have: Ilf*g
llL'
< Ilf
llL1
llg llL1 .
One often summarizes the content of Corollary 3 by saying that L1 is a conwolution algebra. Let A be an arbitrary subset of Rn, and f and g two continuous functions in Rn, one of which has a compact support. Let fA be the
Chap. 26-41
28 1
CONVOLUTION OF FUNCTIONS
function equal to f in A - s u p p g and to zero everywhere else. Then we have obviously, for all x E A,
We may summarize this as follows:
26.1. Let f , g be continuous functions in Rn,one of which has a compact support. Let A be a subset of R". The values of the convolution f * g in the set A do not depend on the values o f f in the complement of the set PROPOSITION
A
-
suppg
= {x E
Rn;x
= x'
- x"
for some x'
E
A', x" E suppg}.
We have already taken advantage of the fact that, if both f and g have compact support, so does f *g. This may be regarded as a corollary of the following result:
PROPOSITION 26.2. Let f , g be as in Proposition 26.1. W e have (26.9)
supp(f* g) C suppf
+ suppg
(vector sum).
+
Proof. Let x belong to the complement of supp f supp g ; then, for y E supp g, x - y belongs to the complement of supp f , hence
COROLLARY. If both supp f and supp g are compact, supp( f compact.
* g) is also
Indeed, if K and H are compact subsets of a HausdorfT TVS E (here R"), K H is also a compact subset of E, as it is the image of the product K x H, compact subset of E x E, under the continuous mapping (x, y ) x y. We proceed now to study convolution from the viewpoint of differentiability. Suppose that f and g are two V1functions, one of which has compact support in Rn.Then it follows immediately from Leibniz' rule .for differentiation under the integral sign that
+
- +
282
DUALITY. SPACES OF DISTRIBUTIONS
[Part I1
is a differentiable function (at every point x) and that
( W A f* g ) = ( ( V x J f )
*o"
=f
* ((a/%)g),
i = 1, ...,n.
In fact, combining this with the fact that the convolution of two continuous functions, one of which has compact support, is a continuous function, we see that f * g is a W function in R". Furthermore, if we apply (26.8) with af/axi instead off, we derive the fact that, for all x E A,
where B = A - supp g. Of course, the right-hand side of the above inequality may very well be infinite. But it will be finite whenever supp g and A are both compact sets. This implies easily the following result, which is a particular case of a more general fact: PROPOSITION 26.3. Let m be an integer, 0 compact support, the convolution
f
< m < +a.If g EL^
has
-f*g
is a continuous linear map of Vm(Rn)into itself.
COROLLARY. Let m, g be as in Proposition 26.3. The convolution f is a continuous linear map of VF(Rn)into itself.
- *g f
Proof of Corollary. In view of the general properties of LF-spaces, it suffices to prove that, for every compact subset K of Rn, the restriction of the mapping f f * g to V T ( K )is continuous, as a map of %?F(K) into VT(Rn)(see Proposition 13.1). But in view of Proposition 26.2, it maps V T ( K ) into VT(K supp g), and on both these spaces the topology induced by Vr(R") is the same as the one induced by P ( R n ) . The corollary follows then directly from Proposition 26.3.
-
+
Exercises 26.1. Let p be a Radon measure, and f a continuous function with compact support in R". Prove that (P
is a continuous function in R".
*f)(x)
=
s
f(x - Y ) M Y )
Chap. 26-61
CONVOLUTION OF FUNCTIONS
283
26.2. Let f,g be two locally L' functions in the real line which vanish identically in the negative open half-line { t E R'; t < O}. Prove that
(f*g)(x)
=
J'f(x
- t ) g ( t ) dt
is a locally L1function in R' vanishing for x < 0. 26.3. Compute the Fourier transform off * g when f,g E Y(R"). 26.4. Let j,g be two continuous functions in R",one of which has compact support. Prove Leibniz' formula:
<
where p , q E N" and q p means q j < pi for all j = 1 ,..., n. 26.5. Let .f be a continuous function with compact support in R". Prove that, if g is exp(
-