Openness and weak openness of multiplication in the space of functions of bounded variation

J. Math. Anal. Appl. 448 (2017) 1560–1571

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Openness and weak openness of multiplication in the space of functions of bounded variation Stanisław Kowalczyk ∗ , Małgorzata Turowska Institute of Mathematics, Pomeranian University in Słupsk, ul. Arciszewskiego 22d, 76-200 Słupsk, Poland

a r t i c l e

i n f o

Article history: Received 2 February 2016 Available online 2 December 2016 Submitted by B.S. Thomson Keywords: Open mapping Weakly open mapping Space of continuous functions of bounded variation Multiplication in function spaces

a b s t r a c t Let BV [0, 1] and CBV [0, 1] be spaces of functions of bounded variation and continuous functions of bounded variation, respectively, with the norm f BV = |f (0)| + V01 (f ). We shall show that the multiplication is a weakly open operation in BV [0, 1] and CBV [0, 1]. Moreover, the multiplication is an open operation in BV [0, 1] with the standard supremum norm. Some other properties of multiplication in function spaces are studied. © 2016 Published by Elsevier Inc.

1. Preliminaries Let C[0, 1] be the space of all continuous real-valued functions defined on [0, 1] with the supremum norm f  = sup |f (t)|. There are some natural operations on C[0, 1], for example, addition, multiplication, t∈[0,1]

minimum and maximum. In [3,9,10] such operations were investigated. All the operations are continuous but only addition, minimum and maximum are open as mappings from C[0, 1] × C[0, 1] to C[0, 1]. Remark 1.1 (Fremlin’s example). [3] In 2004, D.H. Fremlin observed that for f : [0, 1] → R, f (x) = x − 12 , one has f 2 ∈ B 2 (f, 12 ) \ Int B 2 (f, 12 ). Hence multiplication is not an open mapping from C[0, 1] × C[0, 1] into C[0, 1]. Definition 1.2. [3,2] A map between topological spaces is weakly open if the image of every non-empty open set has a non-empty interior. * Corresponding author. E-mail addresses: [email protected] (S. Kowalczyk), [email protected] (M. Turowska). http://dx.doi.org/10.1016/j.jmaa.2016.12.001 0022-247X/© 2016 Published by Elsevier Inc.

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In [3] it is shown that the multiplication in C[0, 1] is a weakly open operation. This was generalized in [7] for C(0, 1) and in [2] for C(X), where X is an arbitrary interval. In [6,7] there are considered some properties of multiplication and others operations in the algebra C(X) of real-valued continuous functions defined on a compact topological space X. Properties of the product of open bals and of n open sets in the space of continuous functions on [0, 1] are studied in [4] and [5], respectively. There is an increasing interest in the study of concepts related to the openness and weak openness of natural bilinear maps on certain function spaces. The reason of it may be that the classical Banach open mapping principle is not true for bilinear maps. In [1], the authors show that multiplication from Lp (X) × Lq (X) onto L1 (X), where (X, μ) is an arbitrary measure space and p1 + 1q = 1, 1 ≤ p, q ≤ ∞, is an open mapping. In the paper we study problem of openness and week openness in the spaces BV [0, 1] of functions of bounded variation and in the space CBV [0, 1] of continuous functions of bounded variation, both defined on [0, 1]. There are two natural norms in BV [0, 1] and CBV [0, 1]: the supremum norm f  = sup |f (t)| and the norm defined by variation f BV = |f (0)| + V01 (f ), where Vab (f ) =

sup

n 

t∈[0,1]

|f (ti ) − f (ti−1 )|

a=t0
is the variation of f on [a, b]. It is worth to mention that (BV [0, 1],  BV ) and (CBV [0, 1],  BV ) are complete, whereas (BV,  ) and (CBV,  ) are not. We use standard notations, N and R denote the set of all positive integers and the set of all real numbers, respectively. B(f, r) and BBV (f, r) denote an open ball centered at f and with the radius r in BV [0, 1] or CBV [0, 1] with the norm   and  BV , respectively. 2. Multiplication in CBV [0, 1] In this section we prove that multiplication in (CBV [0, 1],  BV ) and in (CBV [0, 1],  ) is weakly open. Since Fremlin’s example still works in (CBV [0, 1],  BV ) and in (CBV [0, 1],  ), we know that multiplication is not open neither in (CBV [0, 1],  BV ) nor in (CBV [0, 1],  ). Lemma 2.1. Let f ∈ CBV [0, 1]. For every ε > 0 there exist b > 0, n ∈ N and intervals [u1 , v1 ], [u2 , v2 ], n  . . . , [un , vn ] such that 0 = u1 < v1 < u2 < . . . < vn−1 < un < vn = 1, |f (t)| ≥ b for t ∈ [0, 1] \ [ui , vi ] and n  i=1

i=1

Vuvii (f )

< ε.

Proof. Fix ε > 0. Let {Ik : k ∈ K}, where K is at most countable, be the family of all connected components of the open set [0, 1] \ {t ∈ [0, 1] : f (x) = 0} and ak , bk be the endpoints of Ik . We claim that V01 (f ) =



Vabkk (f ).

(1)

k∈K

The inequality

 k∈K

Vabkk (f ) ≤ V01 (f ) is obvious. Let 0 = x0 < x1 < . . . < xm = 1 be any partition of [0, 1]

and L = {k ∈ K : xi ∈ Ik for some i ≤ m}. Define a new partition Π : 0 = y0 < y1 < . . . < yl = 1 such that  {yj : j ≤ l} = {xi : i ≤ m} ∪ {ak , bk }. Then k∈L m 

|f (xi ) − f (xi−1 )| ≤

i=1

l  j=1

≤

|f (yj ) − f (yj−1 )| ≤ 

{j : ∃k∈L yj ,yj−1 ∈Ik }

|f (yj ) − f (yj−1 )| ≤

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

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≤



Vabkk (f ) ≤

k∈L



Vabkk (f )

k∈K m 

(I denotes the closure of an interval I). Then V01 (f ) = sup

|f (xi ) − f (xi−1 )| ≤

(xi ) i=1

 k∈K

Vabkk (f ). Thus we have

proved (1).  bk Applying (1), we can find a finite set K1 ⊂ K such that Vak (f ) > V01 (f ) − 2ε . Let Vf : [0, 1] → [0, ∞), k∈K1

Vf (t) = V0t (f ). Since f ∈ CBV [0, 1], Vf is uniformly continuous. Hence we can find δ > 0 such that ε |Vf (t1 ) − Vf (t2 )| < 4 card(K if |t1 − t2 | < δ (card(A) denotes the number of elements of the set A). For 1) k ∈ K1 choose αk , βk ∈ (ak , bk ) such that αk < ak + δ, αk < βk and bk − δ < βk . Then 

Vαβkk (f ) >

k∈K1

Vabkk (f ) −

k∈K1



Moreover, [0, 1] \



 2 card(K1 )ε = Vabkk (f ) − 4 card(K1 )

u2 < . . . < vn = 1 (observe that 0 ∈ /



(αi , βi ) and 1 ∈ /

i∈K1

Finally, let b = min

inf

k∈K1 t∈[αk ,βk ]

> V01 (f ) − ε.

(2)

k∈K1

(αk , βk ) can be represented in the form

k∈K1

ε 2



n 

[ui , vi ] for some n ∈ N and 0 = u1 < v1 <

i=1

(αi , βi )). By (2), we obtain

i∈K1

|f (t)| > 0. Then |f (t)| ≥ b for every t ∈



(αk , βk ) = [0, 1] \

k∈K1

completed the proof. 2

n  i=1 n 

Vuvii (f ) < ε.

[uk , vk ]. This

k=1

Lemma 2.2. For every f, g ∈ CBV [0, 1] and every ε > 0 there exist f, g ∈ CBV [0, 1] and a > 0 such that f − fBV ≤ ε, g − gBV ≤ ε and f2 (t) + g2 (t) ≥ a2 for every t ∈ [0, 1]. Proof. Applying Lemma 2.1 to f and 2ε , we can find b > 0, n ∈ N and intervals [a1 , b1 ], [a2 , b2 ], . . . , [an , bn ] n n   Vabii (f ) < 2ε . such that 0 = a1 < b1 < a2 < . . . < bn−1 < an < bn = 1, f (t) ≥ b for t ∈ [0, 1] \ [ai , bi ] and i=1

i=1

Let  K1 = i ∈ {1, 2, . . . , n} : and c = min

inf



inf

t∈[ai ,bi ]

|f (t)| > 0

ε |f (t)| > 0. Fix a = min{ 12n , 2c , 2b }. Again, applying uniform continuity of Vg : [0, 1] →

i∈K1 t∈[ai ,bi ] Vg (t) = V0t (g)

[0, ∞), we can find δ > 0 such that |Vg (t1 ) − Vg (t2 )| < a if |t1 − t2 | < δ. All the more |g(t1 ) − g(t2 )| < a if |t1 − t2 | < δ. For i ∈ {2, 3, . . . , n − 1} \ K1 , let αi = inf{t ∈ [ai , bi ] : |f (t)| = a} and βi = sup{t ∈ [ai , bi ] : |f (t)| = a}. Then [αi , βi ] ⊂ (ai , bi ) and |f (αi )| = |f (βi )| = a. Let K2 = {i ∈ {2, 3, . . . , n − 1} \ K1 : f (αi ) = f (βi )} and K3 = {i ∈ {2, 3, . . . , n − 1} \ K1 : f (αi ) = −f (βi )}. Clearly, K1 , K2 , K3 are pairwise disjoint and K1 ∪ K2 ∪ K3 ⊃ {2, 3, . . . , n − 1}. For i ∈ K3 choose ui , wi , vi ∈ (ai , bi ) such that ai < ui < αi < wi < vi < βi and vi − ui < δ. Define fi : (ai , bi ) → R for i ∈ {2, 3, . . . , n − 1} in the following way. If i ∈ K1 then fi (t) = f (t) for t ∈ (ai , bi ). If i ∈ K2 then

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

fi (t) =

 f (t),

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for t ∈ (ai , αi ) ∪ (βi , bi ),

f (αi ), for t ∈ [αi , βi ].

And if i ∈ K3 then ⎧ ⎪ ⎪f (t), ⎨ fi (t) =

for t ∈ (ai , αi ) ∪ (βi , bi ),

f (βi ), ⎪ ⎪ ⎩f (α ) + i

for t ∈ [wi , βi ], f (βi )−f (αi ) (t wi −αi

− αi ),

for t ∈ [αi , wi ).

Obviously, each fi is continuous and of bounded variation and Vabii (f − fi ) = Vαβii (f − fi ) ≤ Vαβii (f ) + Vαβii (fi ) ≤ Vabii (f ) + 2a. Now, we define f1 : [0, b1 ) → R and fn : (an , 1] → R. If |f (0)| > a or |f (1)| > a then we define f1 , fn in the same way as fi for i ∈ {2, 3, . . . , n − 1}. So, assume that |f (0)| ≤ a and |f (1)| ≤ a. Let β1 = sup{t ∈ [0, b1 ] : |f (t)| = a} and αn = inf{t ∈ [an , 1] : |f (t)| = a} (remind that |f (b1 )| ≥ b > a and |f (an )| ≥ b > a). Define  f1 (t) =

f (t),

for t ∈ (β1 , b1 ),

f (β1 ),

for t ∈ [0, β1 ],

and fn (t) =

 f (t),

for t ∈ (an , αn ), for t ∈ [αn , 1].

f (αn ),

Obviously, f1 and fn are continuous and of bounded variation and V0b1 (f − f1 ) = V0β1 (f − f1 ) ≤ V0b1 (f ), Va1n (f − fn ) = Vα1n (f − fn ) ≤ Va1n (f ), |f1 (t)| ≥ a for t ∈ [0, b1 ) and |fn (t)| ≥ a for t ∈ (an , 1]. Moreover, |f (0) − f1 (0)| ≤ 2a (in the case of |f (0)| > a we have f (0) = f1 (0)). Now, we may define  f : [0, 1] → R by ⎧ ⎪ ⎪ f (t), ⎪ ⎪ ⎪ ⎪ ⎨ f(t) = fi (t), ⎪ ⎪ ⎪ f1 (t), ⎪ ⎪ ⎪ ⎩f (t), n

for t ∈ [0, 1] \ [0, b1 ) ∪ (an , 1] ∪

 (ai , bi ) ,

n−1  i=2

for t ∈ (ai , bi ),

i ∈ {2, . . . , n − 1},

for t ∈ [0, b1 ), for t ∈ (an , 1].

Clearly, f is continuous, |f(0) − f (0)| ≤ 2a and V01 (f − f) =

n  i=1

Vabii (f − f) <

n 

Vabii (f ) + 2na <

ε 2

+ 2na ≤

ε 2

+

ε 6

= 23 ε.

i=1

Therefore, f ∈ CBV [0, 1] and f − f BV = |f(0) − f (0)| + V01 (f − f ) < 23 ε + 2a < ε. Moreover,

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S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

|f(t)| ≥ a for every



t ∈ [0, 1] \

(αk , wk ).

(3)

k∈K3

Now we begin to construct g : [0, 1] → R. Let K4 = {k ∈ K3 : for t ∈ (uk , vk ), since vk − uk < δ. Let

inf

t∈[uk ,vk ]

|g(t)| < a}. Then g(t) ∈ (−2a, 2a)

⎧  (uk , vk ), g(t), for t ∈ [0, 1] \ ⎪ ⎪ ⎪ k∈K4 ⎪ ⎪  ⎪ ⎨a, (αk , wk ), for t ∈ g(t) = k∈K4 ⎪ ⎪ k) ⎪ g(uk ) + a−g(u k ∈ K4 , ⎪ αk −uk (t − uk ), for t ∈ [uk , αk ), ⎪ ⎪ ⎩ g(vk )−a for t ∈ [wk , vk ), k ∈ K4 . a + vk −wk (t − wk ), Clearly, g is continuous, g(0) = g(0) and V01 (g − g) =



Vuvkk (g − g) ≤

k∈K4



(Vuvkk (g) + Vuvkk ( g )) <

k∈K4



(a + 2 · 3a) < 7an < ε.

k∈K4

Obviously, g(0) = g(0). Therefore, g ∈ CBV [0, 1] and g − gBV < ε. By construction, for every t ∈ [0, 1] we have |f(t)| ≥ a or | g (t)| ≥ a. All the more f2 (t) + g2 (t) ≥ a2 . The proof is completed. 2 Lemma 2.3. Let f, g, h : [0, 1] → R be such that  2 2   f (t) + g 2 (t) + 4f (t)g(t) h(t) − f (t)g(t) > 0

(4)

for every t ∈ [0, 1]. Then the functions F, G : [0, 1] → R,   2 h(t) − f (t)g(t) g(t) , F (t) = f (t) +  2   f 2 (t) + g 2 (t) + 4f (t)g(t) h(t) − f (t)g(t) + f 2 (t) + g 2 (t)

(5)

  2 h(t) − f (t)g(t) f (t) , G(t) = g(t) +  2   f 2 (t) + g 2 (t) + 4f (t)g(t) h(t) − f (t)g(t) + f 2 (t) + g 2 (t)

(6)

and

satisfy the equation F G = h. Proof. The equality F G = h may be proved by direct computation. This is quite simple although laborious and we omit it. But it is worth to mention how one can find formulas for F and G. In [7,8] there is presented method of proving of weak openness of multiplication in some function spaces applying the following fact: if (x, y) ∈ R2 , (x, y) = (0, 0) then multiplication of coefficients of points of the plain restricted to the line segment beginning at (x, y) and ending at (x + y, y + x) is strictly increasing, whereas multiplication restricted to the line segment beginning at (x, y) and ending at (x − y, y − x) is strictly decreasing. (Observe that (x + y, y + x) = (x, y) + ∇Φ(x, y) and (x − y, y − x) = (x, y) − ∇Φ(x, y), where Φ is the multiplication of coefficients of points of the plain and ∇Φ is the gradient of Φ.) Thus given functions f, g, h and t ∈ [0, 1], such that |h(t)−f (t)g(t)| is sufficiently small, we can find a point (α, β) which lies on line segment beginning at (f (t) − g(t), g(t) − f (t)) and ending at (f (t) + g(t), g(t) + f (t)) and such that αβ = h(t). Points lying on this line segment are of the form (f (t) + ug(t), g(t) + uf (t)) for

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

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u ∈ [−1, 1]. Thus the equation (f (t) + ug(t)) · (g(t) + uf (t)) = h(t) should be fulfilled and we obtain a quadratic equation f (t)g(t)u2 + (f 2 (t) + g 2 (t))u + f (t)g(t) − h(t) = 0 (if f (t)g(t) = 0, then we have a linear equation). Solutions of this equations are of the form   √ − f 2 (t) + g 2 (t) − Δ = u1 = 2f (t)g(t)  2   2 f (t) + g 2 (t) − Δ −2 h(t) − f (t)g(t) = √ =√   2f (t)g(t) Δ − (f 2 (t) + g 2 (t)) Δ − f 2 (t) + g 2 (t) and   √ − f 2 (t) + g 2 (t) + Δ = u2 = 2f (t)g(t)  2   Δ − f 2 (t) + g 2 (t) 2 h(t) − f (t)g(t) √ √ = ,  = 2f (t)g(t) Δ + f 2 (t) + g 2 (t) Δ + f 2 (t) + g 2 (t) where Δ is the same as earlier. We take the second formula, because it is valid also in the case f (t)g(t) = 0 lim u2 = 0. Then we obtain and, moreover, if we change the value h(t) then h(t)→f (t)g(t)



  f (t) + u2 g(t) g(t) + u2 f (t) =        2 h(t) − f (t)g(t) g(t) 2 h(t) − f (t)g(t) f (t) = f (t) + √ g(t) + √ = h(t). 2 Δ + f 2 (t) + g 2 (t) Δ + f 2 (t) + g 2 (t)

Lemma 2.4. Let f, g ∈ CBV [0, 1] and a > 0 be such that f 2 (t) +g 2 (t) ≥ a2 for t ∈ [0, 1]. Then for every r > 0 there exists δ > 0 such that BBV (f g, δ) ⊂ BBV (f, r)BBV (g, r), where BBV (f, r)BBV (g, r) = {f1 g1 : f1 ∈ BBV (f, r), g1 ∈ BBV (g, r)}. Proof. Since f, g ∈ CBV [0, 1], f, g are bounded and, moreover,

sup |f (t)| = f  ≤ f BV and t∈[0,1]

sup |g(t)| ≤ gBV . t∈[0,1]

First, we will show that if h ∈ CBV [0, 1] and h − f gBV < fulfilled and, moreover

a2 8

then condition (4) from Lemma 2.3 is

 2   a4  2  f (t) + g 2 (t) + 4f (t)g(t) h(t) − f (t)g(t)  ≥ 2 for t ∈ [0, 1]. Indeed,  2    2  f (t) + g 2 (t) + 4f (t)g(t) h(t) − f (t)g(t)  ≥ 2  ≥ f 2 (t) + g 2 (t) − 4|f (t)g(t)| · |h(t) − f (t)g(t)| ≥  2 2 ≥ f 2 (t) + g 2 (t) − 4|f 2 (t) + g 2 (t))| · a8 = = (f 2 (t) + g 2 (t))(f 2 (t) + g 2 (t) −

a2 2 )

≥ a2 ·

a2 2

=

a4 2

for t ∈ [0, 1]. Let K1 = V01 (f 2 +g 2 ), K2 = V01 ((f 2 +g 2 )2 ) and K3 = V01 (f g). Take h ∈ CBV [0, 1] such that h−f gBV < 2 min{1, a8 }. Then, applying properties of variation, we obtain

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V01 ((h − f g)g) ≤ V01 (h − f g)g + V01 (g)h − f g ≤ 2h − f gBV · gBV , V01 (4f g(h − f g)) ≤ 2h − f gBV · 4f gBV and V01



 (f 2 + g 2 )2 + 4f g(h − f g) ≤   V01 (f 2 + g 2 )2 + 4f g(h − f g) ≤  ≤ 2 inf t∈[0,1] |(f (t)2 + g(t)2 )2 + 4f (t)g(t)(h(t) − f (t)g(t))| ≤

Let u =

K2 + 8K3 K2 + 8h − f gBV · f gBV  ≤ a2 a4 2 2

 (f 2 + g 2 )2 + 4f g(h − f g) + f 2 + g 2 and K4 = sup |u(t)| = t∈[0,1]

    = sup  (f 2 (t) + g 2 (t))2 + 4f (t)g(t)(h(t) − f (t)g(t)) + f 2 (t) + g 2 (t) ≤ t∈[0,1]

≤



(f 2 + g2 )2 + 4f  · g + f 2 + g2 .

Clearly, sup |2(h(t) − f (t)g(t))g(t)| ≤ 2h − f g · g ≤ 2h − f gBV · gBV . t∈[0,1] 2

Finally, if h − f gBV ≤ min{1, a8 } and F is defined by formula (5) in Lemma 2.3 then  V01

(F − f ) =

V01

2(h − f g)g

 (f 2 + g 2 )2 + 4f g(h − f g) + f 2 + g 2

 ≤

V01 (2(h − f g)g) sup |u(t)| + V01 (u) sup |2(h(t) − f (t)g(t))g(t)| t∈[0,1]

≤

t∈[0,1] 2

≤

inf |u(t)|

t∈[0,1]

≤

4h − f gBV · gBV K4 +

K

2 +8K3 a2

 + K1 2h − f gBV · gBV

2

(a2 )

=

= h − f gBV · K5 , 

4gBV K4 +2·gBV ·

K2 +8K3 a2

 +K1

where K5 = . Since |F (0) − f (0)| ≤ h − f gBV 2g a4 a2 , we have proved that 2 there exists K > 0 for which F − f BV < Kh − f gBV if h − f gBV < min{1, a8 } (we may take K = K5 + 2g a2 ). 2 In a similar way we can find M > 0 for which G − gBV ≤ M h − f gBV if h − f gBV < min{1, a8 } and G is defined by formula (6) in Lemma 2.3. r Let δ = and take any h ∈ B(f g, δ). If F and G are defined by (5) and (6) in Lemma 2.3, a2 min{

8

,M,K,1}

then F G = h, F − f BV < r and G − gBV < r. It means that BBV (f g, δ) ⊂ BBV (f, r)BBV (g, r).   Theorem 2.5. Multiplication is weakly open in the space CBV [0, 1],  BV .

2

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Proof. The proof follows directly from Lemma 2.2 and Lemma 2.4. Let U be any open set in (CBV [0, 1],  BV ) × (CBV [0, 1],  BV ). We can find f, g ∈ CBV [0, 1] and r > 0 such that BBV (f, 2r) × BBV (g, 2r) ⊂ U . By Lemma 2.2, there exist f, g ∈ CBV [0, 1] and a > 0 such that f − fBV < r, g − gBV < r and f2 (t) + g2 (t) ≥ a for every t ∈ [0, 1]. By Lemma 2.4, there exists δ > 0 such that BBV (fg, δ) ⊂ BBV (f, r)BBV ( g , r) ⊂ BBV (f, 2r)BBV (g, 2r) ⊂ Φ(U ), where Φ denotes multiplication in CBV [0, 1] × CBV [0, 1]. It means that the multiplication is weakly open in the space (CBV [0, 1],  BV ). 2 Remark 2.6. In [2] there is consider a property called dense weak openness, which is a little bit stronger then weak openness. Namely, a mapping f : X → Y is densely weakly open if f (U ) ⊂ Int f (U ) for every open U ⊂ X. Clearly, every densely weakly open mapping is weakly open and there are weakly open mappings which are not densely weakly open. But, it is easily seen, that every continuous and weakly open mapping is densely weakly open. Since multiplication, in considered in this paper spaces, is continuous, we study only the notion of weak openness. Theorem 2.7. Multiplying is weakly open in the space (CBV [0, 1],  ). Proof. The proof follows directly from Lemma 2.2 and Lemma 2.3. Let U ⊂ (CBV [0, 1],  ) × (CBV [0, 1],  ) be any open set. We can find f, g ∈ CBV [0, 1] and r > 0 such that B(f, 2r) × B(g, 2r) ⊂ U . By Lemma 2.2, there exist f, g ∈ CBV [0, 1] and a > 0 such that f − f ≤ f − fBV < r, g − g ≤ g − gBV < r and f2 (t) + g2 (t) ≥ a2 for every t ∈ [0, 1]. ra2  > 0 and take any h ∈ B(f g, ε). Let F ∈ B(f, r), G ∈ B(g, r) be defined by Let ε = 2 max{f ,g}+1 formulas (5) and (6) from Lemma 2.3, respectively. Then F G = h and         2 h − f g g ≤  f − F  =    2   2 2 2 2  f +g + 4f g h − f g + f + g  ≤ h − f g

2g 2g <ε 2
and         2 h − fg f ≤   g − G =    2   2 2 2 2  f +g + 4f g h − f g + f + g  ≤ h − f g

2f  2f  < ε 2 < r. 2 a a

Therefore B(fg, ε) ⊂ B(f, r)B( g , r) ⊂ B(f, 2r)B(g, 2r) ⊂ Φ(U ). It means that multiplication Φ is weakly open in the space (CBV [0, 1],  ). 2 3. Multiplication in BV [0, 1] Lemma 3.1. For every f ∈ BV [0, 1] and ε > 0 there exist f1 ∈ BV [0, 1] and a > 0 such that f − f1 BV < ε and |f1 (t)| ≥ a for every t ∈ [0, 1].

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

1568

Proof. There exist n ∈ N and a partition 0 = t0 < t1 < . . . < tn = 1 of [0, 1] such that V01 (f ) < n  |f (ti ) − f (ti−1 )| + 8ε . Put i=1

Ai = {t ∈ [ti−1 , ti ] : f (t) = 0 or ( lim− f (u) = 0 and t > ti−1 ) or ( lim+ f (u) = 0 and t < ti )} u→t

u→t

for i ∈ {1, 2, . . . , n} and K = {i ∈ {1, 2, . . . , n} : Ai = ∅}. Next, let αi = inf Ai and βi = sup Ai for i ∈ K. ε If αi > ti−1 and lim f (u) = 0 then we take any ai ∈ (ti−1 , αi ) such that |f (ai )| < 16n , otherwise we u→α− i

ε put ai = αi . Similarly, if βi < ti and lim+ f (u) = 0 then we take any bi ∈ (βi , ti ) such that |f (bi )| < 16n , u→βi  otherwise we put bi = βi . It is clear that there exists b > 0 such that f (t) ≥ b for every t ∈ [0, 1] \ [ai , bi ], i∈K

since either ai = ti−1 or lim f (u) > 0 and either bi = ti or lim+ f (u) > 0. t→bi t→a− i    ε Let f1 (t) = f (t) for t ∈ [0, 1] \ [ai , bi ], f1 (t) = 16n for t ∈ (ai , bi ) and if t ∈ {ai , bi } then i∈K

i∈K

i∈K

ε f1 (t) = f (t) if f (t) = 0 or f1 (t) = 16n if f (t) = 0 (obviously, if ai = bi then (ai , bi ) = ∅). By the definition ε of f1 , there exists a > 0 (namely a = min({b, 16n } ∪ {f (ak ) : f (ak ) = 0, k ∈ K} ∪ {f (bk ) : f (bk ) = 0, k ∈ ε . Moreover, (f − f1 )(t) = 0 for K})) such that |f1 (t)| ≥ a for every t ∈ [0, 1] and |f (0) − f1 (0)| ≤ 16n    ε ε t ∈ [0, 1] \ [ai , bi ], (f − f1 )(t) = f (t) − 16n for t ∈ (ai , bi ) and |f (t) − f1 (t)| ≤ 16n for t ∈ {ai , bi }. i∈K

i∈K

i∈K

i i Fix i ∈ {1, 2, . . . , n}. If i ∈ / K then Vtti−1 (f − f1 ) = 0 ≤ Vtti−1 (f ) − |f (ti ) − f (ti−1 )|. So, let i ∈ K and put ai +bi ε ε ci = 2 . It is easily seen that either lim+ f (u) = 0 or |f (ai )| ≤ 16n . Indeed, if ai = αi then |f (ai )| < 16n

u→ai

at once. On the other hand, if ai = αi , lim+ f (u) = 0 and |f (ai )| > u→ai

ε 16n

ai < αi , which is a contradiction. Thus either lim+ f (u) = 0 or |f (ai )| ≤ If |f (ai )| ≤

ε 16n

u→ai

> 0 then lim f (u) = 0 and u→a− i

ε 16n .

then Vacii (f − f1 ) ≤ Vacii (f ) + |(f (ai ) − f1 (ai )) − (f (ai ) − ≤

Vacii (f )

+

ε 16n

+

2ε 16n

≤

Vacii (f )

+

4ε 16n

ε 16n )|

≤

− |f (ai )|,

while if lim+ f (u) = 0 then u→ai

Vacii (f ) = Vacii (f) + |f (ai )| = Vacii (f − − |(f (ai ) − f1 (ai )) − (f(ai ) −

ε 16n )

+ |f (ai )| ≥ Vacii (f − f1 ) + |f (ai )| −

ε 16n )|

≥ Vacii (f − f1 ) + |f (ai )| −

2ε 16n ,

where f(t) = f (t) for t ∈ (ai , ci ] and f(ai ) = 0. Thus we have proved Vacii (f − f1 ) ≤ Vacii (f ) − |f (ai )| + Therefore i Vtci−1 (f − f1 ) = |(f − f1 )(ai )| + Vacii (f − f1 ) ≤

≤

ε 16n

≤

6ε 16n

+ Vacii (f ) − |f (ai )| + +

i Vtci−1 (f )

i ≤ Vtci−1 (f ) +

3ε 8n

ε 4n

≤

− |f (ai ) − f (ti−1 )| − |f (ai )| ≤

− |f (ti−1 )|.

Similarly, we can show Vctti (f − f1 ) ≤ Vctii (f ) + Thus

3ε 8n

− |f (ti )|.

ε 4n .

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

i i Vtti−1 (f − f1 ) ≤ Vtti−1 (f ) +

≤ V01 (f − f1 ) =

i (f ) Vtti−1

n 

+

3ε 4n

− |f (ti )| − |f (ti−1 )| ≤

3ε 4n

− |f (ti ) − f (ti−1 )|,

i Vtti−1 (f − f1 ) ≤ V01 (f ) −

i=1

n 

|f (ti ) − f (ti−1 )| +

3ε 4

1569

< 78 ε

i=1

and f − f1 BV = V01 (f − f1 ) + |f (0) − f1 (0)| ≤ 78 ε +

ε 16n

< ε.

2

Theorem 3.2. Multiplying is weakly open in the space (BV [0, 1],  BV ). Proof. Let U be any open set in (BV [0, 1],  BV ) × (BV [0, 1],  BV ). Take f, g ∈ BV [0, 1] and r > 0 such that BBV (f, 2r) × BBV (g, 2r) ⊂ U . By Lemma 3.1, there exist f ∈ BV [0, 1] and a > 0 such that f − fBV < r and f2 (t) + g 2 (t) ≥ a for every t ∈ [0, 1]. Repeating arguments from the proof of Lemma 2.4, we can prove analogous lemma for the space (BV,  BV ), i.e. there exists δ > 0 such that BBV (fg, δ) ⊂ BBV (f, r)BBV (g, r) ⊂ BBV (f, 2r)BBV (g, 2r) ⊂ Φ(U ). It means that multiplication Φ is weakly open in the space (BV [0, 1],  BV ).

2

Remark 3.3. Notice that for the space (BV [0, 1],  BV ) Fremlin’s example does not work and we have the following question. Question. Is multiplication in (BV [0, 1],  BV ) an open operation? This question is particularly natural, since it turn out that multiplication is an open operation in (BV [0, 1],  ), which we are to prove. Lemma 3.4. Let f, g ∈ BV [0, 1] and r > 0. Then 1. for every x ∈ [0, 1) there exist ε > 0 and δ > 0 such that for every h ∈ BV [0, 1] satisfying h − f g < ε we can find f ∈ BV [x, x + δ] and g ∈ BV [x, x + δ] for which sup |f (t) − f(t)| < r, sup |g(t) − t∈(x,x+δ)

t∈(x,x+δ)

g(t)| < r and fg|(x,x+δ) = h|(x,x+δ) , 2. for every x ∈ (0, 1] there exist ε > 0 and δ > 0 such that for every h ∈ BV [0, 1] satisfying h −f g < ε we can find f ∈ BV [x−δ, x] and g ∈ BV [x−δ, x] for which sup |f (t)−f(t)| < r, sup |g(t)− g (t)| < r t∈(x−δ,x)

t∈(x−δ,x)

and fg|(x−δ,x) = h|(x−δ,x) , 3. for every x ∈ [0, 1] there exists ε > 0 such that for every h ∈ BV [0, 1] satisfying h − f g < ε we can find u, v ∈ R such that |u − f (x)| < r, |v − g(x)| < r and uv = h(x). Proof. (1) Fix x ∈ [0, 1). Let a = lim+ f (t), b = lim+ g(t), K = sup |f (t)| and M = sup |g(t)|. t→x

t→x

t∈[0,1] r min{1,|a|} and 3

t∈[0,1]

First, consider the case where a = 0. Then, let ε = δ > 0 be such that |a − f (t)| < for every t ∈ (x, x + δ). Take any h ∈ BV [0, 1] such that h − f g < ε. Define f: [x, x + δ] → R |(x,x+δ) = h|(x,x+δ) , and g : [x, x + δ] → R by f(t) = a and g(t) = h(t) a for t ∈ [x, x + δ]. It is clear, that f g   f , g ∈ BV [x, x + δ] and sup |f (t) − f (t)| < r. Finally, r min{1,|a|} 3(M +1)

t∈(x,x+δ)

      h(t) − ag(t)  |h(t) − f (t)g(t)|  h(t)   ≤  − g(t) =  + | g (t) − g(t)| =   a a |a|

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S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

+ for t ∈ (x, x + δ) and therefore

r|a| M 3(M ε 2r |g(t)| · |f (t) − a| +1) < + < |a| |a| |a| 3

sup

|g(t) − g(t)| < r.

t∈(x,x+δ)

In the case, where b = 0 we proceed analogously, replacing f by g, a by b an taking ε = δ > 0 such that |b − g(t)| < r min{1,|b|} for every t ∈ (x, x + δ). 3(K+1)

r min{1,|b|} 3

and

Finally, consider the case where a = b = 0. Take any α ∈ (0, min{1, 4r }) and let ε = r min{1,|α|} and δ > 0 3 be such that |f (t)| < α and |g(t)| < r6 for every t ∈ (x, x + δ). Take any h ∈ BV [0, 1] such that h − f g < ε. Define f: [x, x + δ] → R and g : [x, x + δ] → R by f(t) = α and g(t) = h(t) α for t ∈ [x, x + δ]. It is clear, that fg|(x,x+δ) = h|(x,x+δ) , f, g ∈ BV [x, x + δ]) and sup |f (t) − f(t)| ≤ α + α < r. Moreover, t∈(x,x+δ)

      h(t) − αg(t)  |h(t) − f (t)g(t)|  h(t)   ≤  − g(t) =  + | g (t) − g(t)| =   α α α + for t ∈ (x, x + δ) and hence

r (α + α) ε 2r |g(t)| · |f (t) − α| < + 6 < α α α 3

sup

|g(t) − g(t)| < r. Thus we have proved (1).

t∈(x,x+δ)

(2) Evidently, the proof of (2) is almost the same as the proof of (1) and we omit it. (3) If f (x) = 0 then we can take ε = r|f2(x)| and for every h ∈ BV [0, 1] such that h − f g < ε we put |h(x)−f (x)g(x)| u = f (x) and v = h(x) < r. f (x) . Then, obviously, uv = h(x), |u − f (x)| = 0 and |v − g(x)| = |f (x)| Similarly, if g(x) = 0 then we take ε = r|g(x)| and for every h ∈ BV [0, 1] such that h − f g < ε we put 2 h(x) (x)g(x)| v = g(x) and u = g(x) . Then uv = h(x), |v − g(x)| = 0 and |u − f (x)| = |h(x)−f < r. |g(x)| 2

Finally, if f (x) = g(x) = 0 then we take ε = r2 and for every h ∈ BV [0, 1] such that h − f g < ε we 2h(x) r 2ε put u = 2r and v = 2h(x) r . Then uv = h(x), |u − f (x)| = 2 < r and |v − g(x)| = | r | < r = r. The proof is completed. 2   Theorem 3.5. Multiplication is an open mapping in the space BV [0, 1],   . Proof. Let U be any open set in (BV [0, 1],  ) × (BV [0, 1],  ) and take any (f, g) ∈ U . There exists r > 0 such that B(f, r) × B(g, r) ⊂ U . Applying Lemma 3.4, for every x ∈ [0, 1] we can find εx > 0 and δx > 0 such that for every h ∈ BV [0, 1] satisfying h − f g < εx there exist fx ∈ BV ([x − δx , x + δx ] ∩ [0, 1]) and gx ∈ BV ([x − δx , x + δx ] ∩ [0, 1]) for which sup |f (t) − fx (t)| < r, sup |g(t) − gx (t)| < r t∈(x−δx ,x+δx )∩[0,1]

t∈(x−δx ,x+δx )

and (fx gx )|(x−δx ,x+δx )∩[0,1] = h|(x−δx ,x+δx )∩[0,1] . By the compactness of [0, 1], there exist 0 ≤ x1 < x2 < n  . . . < xn ≤ 1 such that [0, 1] ⊂ (xi − δxi , xi + δxi ). Clearly, we may assume that 0 ∈ (x1 − δx1 , x1 + δx1 ), i=1

(xi , xi + δxi ) ∩ (xi+1 − δxi+1 , xi+1 ) = ∅ for i ∈ {1, 2, . . . , n − 1} and 1 ∈ (xn − δxn , xn + δxn ). For i ∈   {1, 2, . . . , n − 1} choose any yi ∈ (xi , xi + δxi ) ∩ (xi+1 − δxi+1 , xi+1 ) and let ε0 = min εxi : i ∈ {1, 2, . . . , n} . We claim that B(f g, ε0 ) ⊂ B(f, r)B(f, r). Take any h ∈ B(f g, ε0 ). Then h − f g < εxi for i ∈ {1, 2, . . . , n}. Define f, g : [0, 1] → R by ⎧  ⎪ ⎪ ⎨fx1 (t), f(t) = fxi (t), ⎪ ⎪ ⎩f (t), x n

t ∈ [0, y1 ], t ∈ (yi−1 , yi ], t ∈ (yn−1 , 1],

i ∈ {2, 3, . . . , n − 1},

S. Kowalczyk, M. Turowska / J. Math. Anal. Appl. 448 (2017) 1560–1571

⎧ ⎪ ⎪ ⎨gx1 (t), t ∈ [0, y1 ], g(t) = gxi (t), t ∈ (yi−1 , yi ], ⎪ ⎪ ⎩g (t), t ∈ (y xn n−1 , 1].

1571

i ∈ {2, 3, . . . , n − 1},

It is obvious that fg = h, f, g ∈ BV [0, 1], f − f < r and g − g < r. Therefore h ∈ B(f, r)B(g, r) and B(f g, ε0 ) ⊂ B(f, r)B(f, r). This completed the proof. 2 It is worth noting the importance of Lemma 2.3. As far, functions f, g satisfying equality f g = h for given h were obtained by “gluing” functions defined only locally. Quite a few results from previous papers follows from Lemma 2.3, for example Theorem 5 from [3], Theorem 1 from [7] and Lemma 2 from [6]. Moreover, Lemma 2.3 may be easily used for the proof of weak openness of multiplication in some other function spaces, for example in the space of absolutely continuous functions. Theorem 3.6. Multiplying is a weakly open map in the space of absolutely continuous functions with the supremum norm. Proof. It is enough to observe that if f, g, h are absolutely continuous then F, G defined in Lemma 2.3 are absolutely continuous and functions f, g constructed in Lemma 2.2 are absolutely continuous too. 2 References [1] M. Balcerzak, A. Majchrzycki, A. Wachowicz, Openness of multiplication in some function spaces, Taiwanese J. Math. 17 (2013) 1115–1126. [2] M. Balcerzak, A. Maliszewski, On multiplication in spaces of continuous functions, Colloq. Math. 122 (2011) 247–253. [3] M. Balcerzak, A. Wachowicz, W. Wilczyński, Multiplying balls in the spaces of continuous functions on [0, 1], Studia Math. 170 (2005) 203–209. [4] E. Behrends, Walk the dog, or: products of open balls in the space of continuous functions, Funct. Approx. Comment. Math. 44 (2011) 153–164. [5] E. Behrends, Products of n open subsets in the space of continuous functions on [0, 1], Studia Math. 204 (2011) 73–95. [6] A. Komisarski, A connection between multiplication in C(X) and the dimension of X, Fund. Math. 189 (2006) 149–154. [7] S. Kowalczyk, Weak openness of multiplication in C(0, 1), Real Anal. Exchange 35 (2010) 235–241. [8] S. Kowalczyk, On operations in C(X) determined by continuous functions, Acta Math. Hungar. 142 (2013) 56–71. [9] A. Wachowicz, Baire category and standard operations on pairs of continuous functions, Tatra Mt. Math. Publ. 24 (2002) 141–146. [10] A. Wachowicz, On Some Residual Set, PhD dissertation, Łódź Technical Univ., Łódź, 2004 (in Polish).