CC BY
6Probl. Anal. Issues Anal. Vol. 6(24), No. 1, 2017, pp. 11-18
DOI: 10.15393/j3.art.2017.3871
n
UDC 517.518
E. S. Belkina, Yu. V. Malykhin
FOURIER COEFFICIENTS OF CONTINUOUS FUNCTIONS WITH RESPECT TO LOCALIZED HAAR
SYSTEM
Abstract. We construct a nontrivial example of a continuous function f * on [0,1]2 which is orthogonal to tensor products of Haar functions supported on intervals of the same length. This example clarifies the possible behaviour of Fourier coefficients of continuous functions with respect to a localized Haar system. The function f* has fractal structure. We give lower bounds on its smoothness.
Key words: fractals, Haar system, Haar wavelets
2010 Mathematical Subject Classification: 26B35, 42C10
1. Introduction. Let Ak,i = (i2k1, 2k); by A+ and A— we denote the left and right halves of Ak,i, correspondingly.
Define the localized Haar system on [0,1]2 in the following way (see fig. 1): xo(x,y) = 1 on [0,1]2; other functions of the system are grouped into packs; for k = 0,1, 2,... the k-th pack consists of the functions
xk2),j(x,y) = xkj(У,xk3),jy) = 2 kxkijy)xkljy),
where i,j = 1, 2,..., 2k. (The values on the boundary of Ak,i x Ak,j are not important for us.)
©Petrozavodsk State University, 2017
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Fig. 1. Localized Haar system
We should note that the localized Haar system comes from the general construction of multivariate wavelets. Suppose that {Vk} is a multiresolution analysis (MRA; see [2, Sect. 1.2]) in L2 (R) with scaling function ^o ^ Vo. Denote by Wk the orthogonal complement to Vk in Vk+1. MRA generates the wavelet function ^ e W0 and its dyadic dilations and shifts } form a basis of L2(R). There are at least two ways how to construct multivariate wavelets from univariate ones.
• The first way is to take the tensor product of one-dimensional wavelet functions: := ^k^i < ^kk. The disadvantage of this method is that the new system is not "localized" even if the one-dimensional wavelets were: functions with large K may have large support.
• The second way is to take the tensor product of the one-dimensional MRA's: Vk = Vk(1) < Vk(2). Then we define wavelet spaces Wk by Vk+1 = Vk ® Wk and take basis in Wk.
The localized Haar system is obtained using the tensor product of simplest MRA with the scaling function ^ = 1[0,1) and the wavelet function ^ = 1[0,1/2) — 1[1/2,1). Three functions ^(1) = ^ < ^(2) = ^ < ^ and ^(3) = ^ < ^ form a basis of W0. Dyadic dilations and shifts of that functions (inside [0,1]2) produce a localized Haar system, which is an orthonormal basis in L2[0,1]2 (see also [2, Sect. 2.8]).
For any summable function f: [0,1]2 ^ R one can consider its FourierHaar coefficients:
cM,j(f) = ff f(x, y)xkppi,j(x, y) dxdУ, P = 1 2,3. [0,1]2
It was observed by B. I. Golubov (see, e.g., [1, Chapter 3, §2]) that for a classical one-dimensional Haar system {xk,i} and for any f e C[0,1],
f = const, one has
Hm 23k/2 max |cfc,i(f)| > 0,
k^TO i '
1
where ck,i(f) = / f (x)xk,i(x) dx.
o
What happens on the square? If f (x,y) = g(x) + h(y), everything reduces to the one-dimensional case. Indeed, ck^j(f) = 2_k/2ck,i($),
ci2] j (f) = 2-k/2ck ,j (h) and ck3 j (f) = 0. So we shall not consider this trivial situation. In the general case, P. V. Glebov in his diploma work had proven the following.
Theorem A. If f is a continuous function on [0,1]2 and it is not represented as f (x, y) = g(x) + h(y), then
Hm 4k max |cjf) 71 > 0, p = 1, 2.
k^TO i,j , ,j
We reproduce the proof here; it is rather short. Proof. Suppose, f E C[0,1]2 and lim 4k max Ick"1] 7-1 = 0; we shall prove
fc^TO i,j , ,j
that f does not depend on x. Consider the "vertical average" function 1
f ° (x) := f f (x,y) dy. It is easy to compute that
0
ck,i(f °)=2-k/2 di (f). j=1
So, 23k max |ck,i(f °)| ^ 0 and therefore f ° = const.
1
For any dyadic square Ak,i x Ak,j we can analogously consider the "vertical average" function of f on this square and show that it is constant.
If there are x1, x2, y0 E [0,1], x1 = x2, such that f (x1, y0) = f (x2, y0), then
J f (x1,y) dy = J f (x2, y) dy (1)
for some small interval Ak,j 9 y0. But the function v(x) = / f (x,y)dy
is continuous, and we have proven that it is constant on each Ak,i. Hence, it is constant on [0,1], but this contradicts (1). □
So, c(1) - and c(2)-coefficients of continuous functions cannot decay very fast in nontrivial sutiations. What can one say about c(3)-coefficients? The main result of this paper is a rather surprising example of a function with identically zero c(3)-coefficients.
Let us formulate our theorem. By Lip(a) we denote the space of functions f : [0,1]2 ^ R such that the inequality
|f (x, y) - f (x', y')| < C(|x - x'| + |y - y'|)a, (x, y), (x', y') G [0,1]2
holds for some C = C (f ).
Theorem 1. There exists a continuous function f * : [0,1]2 ^ R which is not represented as f*(x, y) = g(x) + h(y), such that C3 j(f *) = 0 for all k,i,j. Moreover, f * lies in Lip(0.64).
We do not know the maximum possible smoothness of such functions. In particular, it would be interesting to construct a Lip(1) example of f *, or to prove that it does not exist. One can see that there are no "simple" examples:
• the c(3)-coefficients of a C2-smooth function f (x,y) = g(x) + h(y) cannot decay as o(8-k):
IïmSfc max |cg,j(f)| > 0; (2)
• the c(3)-coefficients of a piecewise-linear1 function f (x,y) = g(x) + +h(y) cannot decay as o(4-k).
d 2 f
Let us prove this. If f G C2[0,1]2, the^ = 0 in some point
dxdy
d 2f
(otherwise f (x,y) is represented as g(x) + h(y)). So,
dxdy
> co > 0 in
d 2 f
some neighbourhood of that point. Using that ck3) (f) = 2-3k-4 ■ 0 (f)
k,l,J dxdy
for some f e Ak,^ x Ak,j, we obtain (2).
Consider a piecewise-linear function: the square [0,1]2 is triangulated, f is linear on each triangle. One can show that there exists two triangles ABC and ABD, having a common side AB, which is not vertical nor
1 We say that a function f : [0, 1]2 ^ R is piecewise linear, if f is continuous on [0, 1]2 and for some triangulation of the square, f is linear on each triangle.
horizontal, such that the linear functions corresponding to ABC and ABD differ. We can subtract a linear function from f (this does not change c(3)) and make it equal to zero on the triangle ABC. Moreover, we normalize f to ensure that (df/dx)2 + (df/dy)2 = 1 on ABD. For any sufficiently large k there exists a dyadic square Ak,i x Ak,j, which intersects AB and does not intersect any triangles other that ABC, ABD. One can prove that ck3)j(f) = 0 in this "standard" situation2. It follows (using the
compactness argument and scaling) that max |ck3) 7-(f)| > R■ 4-k for large
i,j k,i,j
k, where the constant R depends only on the slope of the segment AB. The next two sections are devoted to the proof of our theorem. 2. Construction of the function f *.
Proof of Theorem 1. We consider linear functions
L(x, y) = Ax + By + C such that J J L(x, y) dx dy = 0. (3)
[0,1]2
It is clear that the function L(x, y) from (3) with L(0, 0) = a, L(0,1) = 6, L(1,1) = c, L(1, 0) = d exists if and only if a + c = 6 + d = 0. We denote such (unique) function as La,b; it is just La,b(x,y) = a — (a + 6)x — (a — 6)y. These functions will be the "building blocks" of our construction. Note that always ck,),j(La,b) = 0.
For 6 E R, 6 = 1, we consider the space Cb of functions f: [0,1]2 ^ R with the following properties:
(i) f is continuous on [0,1]2;
(ii) f (0, 0) = 1, f (0,1) = 6, f (1,1) = 1, f (1, 0) = 6;
(iii) f is linear on the sides of [0,1]2;
(iv) // f(x,y) dxdy = °. [0,1]2
Suppose that for some t, Ai, ui E R (i = 1, 2, 3, 4) the following equalities hold:
1,^= A1 (1,6,1,6)+A2(6,1,6,1) + (u1 ,U2, — U1, -U2), (4)
2It is sufficient to prove that co,i,i{/) = 0 for f (x,y) = max(0, ax + by + c) if line ax + by + c = 0 intersects the interior of [0,1]2. This can be done by considering different cases of position of line ax + by + c = 0.
b +1 b +1
, b, ——,t) = A3(1,b, 1,b)+A4(b, 1,b, 1) + (u3,U4, — «3, —«4). (5)
22
Then we can define the operator T: Cb ^ Cb by the following rules: Tf (X, 2) = Tf (1 — X, 1 — 2) = A1 f (x,y) + A2f (1 — x,y) + LU1 ,U2(x,y),
Tf (x, ^) = Tf (1—2, v) =
= A3f (x, y) + A4f (1 — x, y) + LU3,U4 (x, y).
Let us show that Tf indeed lies in Cb. First, note that properties (ii) and (iii) of Cb imply that f (0,1/2) = f (1/2,1) = f (1,1/2) = f (1/2, 0) = = (b + 1)/2 and the same should hold for Tf. Consider the definition of Tf on the square [0,1/2]2. The equality (4) implies that
(Tf (0, 0), Tf (0,1/2), Tf (1/2,1/2), Tf (1/2, 0)) = (1, ^
'(6)
In particular, t = Tf(1/2,1/2). Definition for other quater-squares is similar. Note that the values of Tf on common vertices of that squares agree. Now we check properties (i)-(iv) of Tf: (ii) follows from (6) and analogous equalities; properites (iii) and (iv) follow from the corresponding properties of f and La,b. Finally, Tf is continuous because values of Tf on different quarter-squares are continuously "glued" by linearity of Tf on the sides of all that squares. Soon we will show that if
k := max(|A11 + |A21, A | + |A41) < 1,
then T is a contraction (say, in the uniform metric). We claim that one can take f * to be the fixed point of T. Indeed, f * is not represented as g(x) + h(y) because f (0, 0) + f (1,1) = f (0,1) + f (1, 0) (recall that
b = 1). Next, f* is continuous by construction. Finally, we check that
(3)
ckij(f*) = 0 using induction on k. If k = 0, it follows from the fact that the average of f * on [0,1]2 is zero and hence it is zero on all quarter-squares. Suppose that ck3-1 i j (f *) = 0 for all i, j. Note that ck3 j (La , b) = = 0. So, if 1 < i, j < 2k-1, we have
cg, j (f *) = cg, j (Af * (2x, 2y) + A2f * (1 — 2x, 2y) + L^, „2 (2x, 2y)) =
= 1 (A1ck3-1,i,j (f *) + A2 ck3)1,2k-1 -i+1,j (f *)) = 0.
Other cases are analogous.
Here is a proof-of-concept simple example of configuration for T with
k=3/4 6=o, t=- i, (Ai)=(4,2,2, - i), («i)=(4,0,0, {
3. Contractivity of T. Recall that we want to prove more than just the continuity of f*, but also some smoothness. Therefore we will work in the space Cb if Lip(a) and find a fixed point f * of T there. We need a convenient metric to prove contractivity of T. Denote by Sk,i,j the square Ak,i x Ak,j, and, as usual, w(f, S) = sup |f (A) — f (B)|. It is
rather clear that a continuous functions belong to Lip(a) if and only if
max w(f, Sk,i,j) = O(2-ka). So, we take
1<i,j<2fe
P(f,g) = sup2ka maX W(f — g,Sk,i,j ).
k>0 1<i,J<2fe
It is indeed a complete metric in Cb if Lip(a). (Note that if f — g = const, f, g E Cb, then f = g.) Also, T preserves space Cb if Lip(a), as one can see, e.g., from (8). Let us prove that
p(Tf,Tg) < k 2ap(f,g), (7)
hence T is a contraction provided k < 2-a. We should estimate
w(Tf — Tg,Sk,i,j). Suppose that Sk,i,j C [0,1/2]2. Then, by definition of T,
w(Tf — Tg,Sk,i,j) = = w(A1 f (2x, 2y) + A2f (1 — 2x, 2y) — A1 g(2x, 2y) — A2g(1 — 2x, 2y), Sk,i,j) < < |A1 |W(f — g, Sk-1,i,j ) + | A2 Kf — g, Sk-1,2k-i-i+1,j ) <
< K2-k-1)Xf,g). (8)
Other cases are completely analogous. The claim (7) follows.
The last thing to do is to minimize k. One should use the following configuration (t and ui may be recovered using (4, 5)):
. —5 + -^17 1 —1 + -^17 —1 — -^17
6 = -, A1 = -, A2 = -, A3 = 0, A4 =-.
2 11 4' 2 8 ' 3 ' 4 8
This gives k = (1 + v/l7)/8 = 0.6403 ... and a < - log2 k = 0.6429 .... The theorem is proven. □
Fig. 2. Function f * References
[1] Kashin B. S., Sahakyan A. A. Orthogonal series. American Mathematical Soc., 2005, 451 p.
[2] Novikov I.Ya., Protasov V.Yu., Skopina M. A. Wavelet theory. American Mathematical Soc., 2011, 506 p.
Received April 15, 2017. In revised form, June 15, 2017. Accepted June 19, 2017. Published online June 21, 2017.
Petrozavodsk State University
33, Lenina pr., Petrozavodsk 185910, Russia
E-mail: [email protected]
Steklov Mathematical Institute 8, Gubkina st., Moscow 119991, Russia E-mail: [email protected]
