THE BANG SOLUTION OF THE COEFFICIENT PROBLEM – 1 Possible introduction (written in common English).

They call themselves Analysts, but what they do

is just manipulating inequalities !... (ascribed to Arnold)

Let T be a measure space with probability measure μ. Let ψ_{j} : T → ℝ be an orthonormal system of functions in L^{2}(T), i.e. ||ψ_{j}||_{2} = 1 for every j and (ψ_{i}, ψ_{j}) =^{def} ∫_{T} ψ_{i}ψ_{j}dμ = 0 for every i ≠ j. Suppose that X is some other space of functions on T such that ||f||_{2} ≤ ||f||_{X} for every f ∈ X (the main example to keep in mind is X = L^{∞}(T)). The question we are going to discuss is whether we can say anything nontrivial about the decay of the Fourier coefficients (f, ψ_{j}) = ∫_{T} fψ_{j}dμ of functions f ∈ X.

Of course, we always have the Bessel inequality:

∑_{j}(f, ψ_{j})^{2} ≤ ||f||_{2}^{2} ≤ ||f||_{X}^{2}.

\underline{Definition}

We will call the space X large (with respect to the system {ψ_{j}}) if for every sequence {a_{j}} of positive numbers satisfying ∑_{j} a_{j}^{2} = 1 there exists a function F ∈ X such that |(F, ψ_{j})| ≥ a_{j} for every j.

Roughly speaking, this definition means that the Bessel inequality is the only thing about the decay we may say for sure.

Note that if X is large, we can find a solution F of the coefficient problem |(F, ψ_{j})| ≥ a_{j} with uniformly bounded norm (i.e. such that ||F ||_{X} ≤ C where C > 0 does not depend on a_{j}). Indeed, otherwise for every k = 1, 2, … one can choose a sequence {a_{j}^{(k)}} with ∑_{j}(a_{j}^{(k)})^{2} = 1 and such that there is no function F ∈ X with ||F ||_{X} ≤ 4^{k} satisfying

(F, ψ_{j})| ≥ a_{j}^{(k)} for all j. But then for the sequence a_{j} =^{def} ∑_{k=0}^{∞} 2^{−k}a_{j}^{(k)} there is no F ∈ X solving the coefficient problem |(F, ψ_{j})| ≥ a_{j} (because a_{j} ≥ 2^{−k}a_{j}^{(k)} for every k and thereby we should have ||F ||_{X} ≥ 2^{k} for every k which is impossible), but still ∑_{j} a_{j}^{2} ≤ 1.

Which spaces are large and which are not ? There exists the following

\underline{Conjecture:}

X is large iff there is a constant c > 0 such that for every sequence {b_{j}} of positive numbers satisfying ∑_{j} b_{j}^{2} = 1 one can find f ∈ X such that

||f||_{X} = 1 and ∑_{j} b_{j}|(f, ψ_{j})| ≥ c.

The conjecture means that the set of sequences { {|(f, ψ_{j})|} : f ∈ X, ||f||_{X} ≤ 1} (which is star-like with respect to the origin, but by no means convex) can be treated up to a certain extent as a convex set and one may switch from the problem to its dual and back.

In most cases when X is known to be large (including those we will add below to the existing list) it is very easy to see that the element f solving this dual problem exists. An interesting exception is the space X = U(T) of the functions defined on the unit circumference 𝕋 with uniform convergent Fourier series with respect to the standard base {z^{k}}_{k=−∞}^{∞} (the norm in U(𝕋) is defined by ||f||_{U}(𝕋) =^{def} sup_{n,m∈ℤ} ||S_{n,m}f(z)||_{∞} where

S_{n,m}f(z) =^{def} ∑_{k=n}^{m} f̂(k)z^{k}). It is known that U(T) is large with respect to any orthonormal system {ψ_{j}} satisfying sup_{j} ||ψ_{j}||_{∞} < +∞, say, but to check solvability of the dual problem for this space is only a very little bit easier then to solve the coefficient problem itself.

Despite the conjecture above is quite old, there has been found no evidence of that the fact so general should hold. So, a lot of people now (me among them) believe that it is false. Meanwhile de-Leeuw, Kahane, Katznelson and others invented quite powerful technique allowing to check that X is large for many particular spaces X. Their idea was to consider the random sum f_{ε} =^{def} ∑_{j} ε_{j}a_{j}ψ_{j} (where ε_{j} = 1 or ε_{j} = −1 with probability \frac{1}{2} each) and to show that with positive probability this sum “almost belongs” to X. The exact meaning of this phrase is that for every A > 0 it is possible to decompose f_{ε} into the sum f_{ε} = g_{A} + h_{A} such that ||g_{A}||_{X} ≤ A, ||h_{A}||_{2} ≤ Δ(A) and Δ(A) decays fast enough as A → +∞. Then they proceeded by “minor corrections” as follows:

Suppose that we have a function F ∈ X of norm ||F ||_{X} ≤ M which solves the coefficient problem for all j but those in some exceptional set J and that ∑_{j∈J} a_{j}^{2} ≤ b^{2} (for instance, we may start with F ≡ 0, M = 0, J which is the whole index set and b = 1). Let us choose the signs ε_{j} in such a way that f_{ε} = ∑_{j∈J} ε_{j}a_{j}ψ_{j} almost belongs

to X. Note that the function F̃ =^{def} F + 2f_{ε} solves the coefficient problem |(F̃, ψ_{j})| ≥ a_{j} for every j but, unfortunately, does not belong to X in general. Nevertheless, it lies not very far from X: as for every A > 0 we can decompose f_{ε} into the sum g_{A} + h_{A} where ||g_{A}||_{X} ≤ Ab and ||h_{A}||_{2} ≤ Δ(A)b (the factor b is due to the condition ∑_{j∈J} a_{j}^{2} ≤ b^{2}, not 1, as in the definition of Δ(A) above), the functuon F̃ can be decomposed into the sum of F_{A} =^{def} F + g_{A} and h_{A}. When we take F_{A} instead of F̃, some of the Fourier coefficients get spoiled but since the L^{2}-norm of the difference h_{A} is small, we cannot substantially spoil too many of them. Namely, let c ∈ (0, 1) and let J′ = {j : |(F_{A}, ψ_{j})| < (1 − c)a_{j}}. Then for every j ∈ J′ we have |(h_{A}, ψ_{j})| ≥ ca_{j} and therefore

∑_{j∈J′} a_{j}^{2} ≤ c^{−2}||h_{A}||_{2}^{2} ≤ \frac{Δ(A)^{2}b^{2}}{c^{2}} =^{def} b′^{2}.

For j ∉ J′ the coefficients (F_{A}, ψ_{j}) are easy to repair: just take F′ =^{def} \frac{1}{1−c}F_{A}. Thus, having started with the function F ∈ X of norm ||F ||_{X} ≤ M and with the exceptional set J for which ∑_{j∈J} a_{j}^{2} ≤ b^{2}, we got a new function F′ of norm ||F′||_{X} ≤ M′ =^{def} \frac{M+Ab}{1−c} and a new exceptional set J′ for which ∑_{j∈J′} a_{j}^{2} ≤ b′^{2} where b′ = \frac{Δ(A)b}{c}. If we contrive to get b′ < b without essential increase in the norm estimate M, we will be able to iterate the procedure. To have the process convergent, we need to construct three sequences {A_{k}}, \{b_{k}\} and \{c_{k}\} (k = 0, 1, …), for which b_{0} = 1; b_{k+1} = \frac{Δ(A_{k})b_{k}}{c_{k}}, c_{k} ∈ (0, 1) for every k; \{b_{k}\} decreases to 0 as k → ∞ and, at last, both series ∑_{k=0}^{∞} c_{k} and ∑_{k=0}^{∞} A_{k}b_{k} converge (the last condition comes just from the necessity of having an uniform bound for the norms of the corresponding functions F_{k} ∈ X, but what it actually implies is ∑_{k} ||F_{k} − F_{k+1}||_{X} < +∞ which is more than enough for the existence of the limit F = lim_{k→∞} F_{k} ∈ X).

Let’s now figure out when such sequences exist. As c_{k} = \frac{Δ(A_{k})b_{k}}{b_{k+1}} , the problem re-

duces to the choice of \{b_{k}\} and \{A_{k}\} satisfying the conditions ∑_{k=0}^{∞} \frac{Δ(A_{k})b_{k}}{b_{k+1}} < +∞ and ∑_{k=0}^{∞} A_{k}b_{k} < +∞. (If we take just arbitrary sequences satisfying these conditions, several first c_{k} may turn out to be greater than 1, but this can be easily corrected by enlarging finite number of corresponding A_{k}’s). Uniting these two conditions in one

∑_{k=0}^{∞} b_{k} (\frac{Δ(A_{k})}{b_{k+1}} + A_{k}) < +∞,

we see that we also have almost no freedom in choosing {A_{k}} if {b_{k}} is given: A_{k} should minimize the expression \frac{Δ(A)}{b_{k+1}} +A. Since Δ(A) is a decreasing function and since we do not

care about constant factors like 2, the best choice comes from solving \frac{Δ(A)}{b_{k+1}} = A. Denoting the solution of the equation Δ(A) = At by A(t) (for any continuous decreasing function Δ it exists for all t > 0 and is a decreasing continuous function of t), we find out that all we need is to construct a sequence {b_{k}} starting with b_{0} = 1 and decreasing to 0 such that ∑_{k} b_{k}A(b_{k+1}) < +∞.

The trivial estimate

∑_{k} b_{k}A(b_{k+1}) > ∑_{k}(b_{k} − b_{k+1})A(b_{k+1}) ≥ ∫_{0}^{1} A(t)dt

shows that we should assume at least that ∫_{0} A(t)dt < +∞. On the other hand, whatever Δ is, A(t) turns out to be quite regular function. Indeed, since \frac{t}{2}(2A(t)) = tA(t) = Δ(A(t)) ≥ Δ(A(\frac{t}{2})), we have A(\frac{t}{2}) ≤ 2A(t) for every t > 0.

Thus, taking b_{k} = 2^{−k}, we get

∑_{k=0}^{∞} b_{k}A(b_{k+1}) = ∑_{k=0}^{∞} 2^{−k}A(2^{−k−1}) ≤ 4 ∫_{0}^{1} A(t)dt.

So, the condition ∫_{0} A(t)dt < +∞ is also sufficient.

What does it mean in terms of the original function Δ ? Changing variable from t to A, we get ∫^{∞} Ad (\frac{Δ(A)}{A}) < +∞ and it remains only to integrate by parts to obtain the celebratred

De-Leeuw – Kahane – Katznelson theorem:

If ∫ ^{∞} \frac{Δ(A)}{A}dA < +∞, then X is large in L^{2}(T) with respect to the system ψ_{j}.

For instance, when X = L^{∞}(T) and the norms ||ψ_{j}||_{∞} are uniformly bounded, we may use the Khinchin theorem to get that f_{ε} ∈ L^{4}, say, with estimate for the norm ||f_{ε}||_{4} not depending on the sequence {a_{j}}. This gives Δ(A) ≤ CA^{−1} which is more than enough to apply the theorem and to conclude that L^{∞}(T) is large.

Though quite powerful in general, the de-Leeuw – Kahane – Katznelson approach has two unpleasant restrictions: to proceed their way one should assume at the very least that 1) X is dense in L^{2}(T);

2) All ψ_{j} almost belong to X themselves.

Correspondently, there remained two problems which could not be solved by the de-Leeuw – Kahane – Katznelson approach:

1. The ”support” problem:

Let T′ be a subset of T of positive measure. Is the space X = L^{p}(T′) =^{def} {f ∈ L^{p}(T) : supp f ⊂ T′} (2 ≤ p ≤ ∞) large or not ?

and

2. The “minimal assumptions” problem:

Let again X = L^{p}(T). What are the minimal requirements on the system {ψ_{j}} which garantee that X is large with respect to this system ?

The first question was open even for p = 2, T = 𝕋, (as usual, we denote by 𝕋 the unit circumference with the Haar measure on it) and the standard base ψ_{j} = z^{j} (j ∈ ℤ)! Of course, the density of X in L^{2}(T) can be restored if we completely forget about T \ T′ and consider T′ as the whole space (with the measure dμ′ =^{def} \frac{dμ}{μ(T′)}). Then we should also

forget about the the original system ψ_{j} and consider the functions ψ′_{j} =^{def} \sqrt{μ(T′)}ψ_{j}|_{E}. (the renormalization which preserves L^{2}-norm) instead. Unfortunately the system {ψ′_{j}} is no longer orthogonal. In general it can be just an arbitrary system of functions satisfying

|| ∑_{j} c_{j}ψ′_{j}||_{L^{2}(T′,μ′)} ≤ (∑_{j} c_{j}^{2})^{\frac{1}{2}} . And it is also fatal: first, now we cannot say that the coefficients (f_{ε}, ψ′_{j}) of the sum f_{ε} = ∑_{j} ε_{j}a_{j}ψ_{j} are just ε_{j}a_{j} and second, we no longer have a possibility to correct one coefficient without substantial influencing others (just because ψ′_{j} may be even linearly dependent now) and it seems that the entire idea of “minor corrections” fails.

As to the second problem, the de-Leeuw – Kahane – Katznelson technique can be applied only if we assume a priori something like ψ_{j} ∈ L^{2} log^{2+δ} L, but there is no evident reason for a condition of this kind to be necessary (and actually it is not).

What are we going to do ?

Consider the sets S_{j} =^{def} {f ∈ L^{2}(T) : |(f, ψ_{j})| < a_{j}}. Obviously, X is large iff the unit ball B_{X} =^{def} {f ∈ X : ||f||_{X} ≤ 1} of the space X cannot be covered by sets S_{j} with arbitrarily small ∑_{j} a_{j}^{2}. Since in general B_{X} is just a convex set, symmetric with respect to the origin and since from the geometric point of view S_{j} is a strip of width 2a_{j}, the whole thing resembles a lot the famous “plank problem” posed by Tarski in the very beginning of the century and solved by Bang in 1950s:

Plank problem:

Given a convex set B ∈ ℝ^{2} (or ℝ^{n}, for Bang’s solution it is all the same), is it possible to cover it by several strips of total width less than the width of B ? (The width of a convex body is defined as the width of the narrowest strip containing it).

The conjectured answer was “no”, but it took about 40 years to prove that. The proof, when found, was … 2 pages long and required from the reader only basic knowlege of elementary geometry. For reader’s convenience it is included into this paper as an appendix.

Of course, to solve the coefficient problem as it was stated above you cannot just apply the result (because the width of B_{X} is 0 unless X = L^{2}(T) and because we need to estimate from below the sum ∑ a_{j}^{2}, not ∑ a_{j}), but it turns out that a minor modification of the proof is enough. (So minor that I actually even do not pretend to be an author of the next two sections; rather I act there like a shadow that enters and goes over many strange places which completely eliminate the attention of his master just passing by).

We will deal with both the “support” and the “minimal assumptions” problems simultaneously, so, from the very beginning we will assume that the inequality

∥∑_{j} c_{j}ψ_{j}∥_{2} ≤ (∑_{j} c_{j}^{2})^{\frac{1}{2}}

is the only a priori information about the system ψ_{j} we have.

THE BANG SOLUTION OF THE COEFFICIENT PROBLEM - 2 Between L^{2} and L^{∞} (written in Analysis language). Let T be a measure space with probability measure μ. Let, as before, {ψ_{j}} be a

system of functions satisfying || ∑_{j} c_{j}ψ_{j}||_{2} ≤ (∑ c_{j}^{2})^{\frac{1}{2}} for every c_{j} ∈ ℝ.

The question we are going to solve below is when X = L^{p}(T) (2 ≤ p ≤ ∞) is large with respect to the system {ψ_{j}}. Also we will give a good estimate for the norm of the function F, solving the coefficient problem.

Trivial necessary condition.

In general it is ||ψ_{j}||_{X^{∗}} ≥ β > 0. For the case X = L^{p}(T) this means
(∫_{T} |ψ_{j}|^{q})^{\frac{1}{q}} ≥ β > 0  for every j  (B_{p})

where q is the exponent conjugate to p, i.e. \frac{1}{p} + \frac{1}{q} = 1.

Our aim is to show that (B_{p}) is also sufficient. More precisely, we have the following

\underline{Theorem:}

If the system {ψ_{j}} satisfies (B_{p}), then for every sequence of positive numbers a_{j} satisfying ∑_{j} a_{j}^{2} = 1 we can find a function F ∈ L^{p}(T), such that

||F ||_{p} ≤ (\frac{3π}{2} )^{1− \frac{2}{p}} β^{−2} ≤ 5β^{−2} and |∫_{T} F ψ_{j}dμ| ≥ a_{j} for every j.

\underline{Remark:}

1) Without additional assumptions on ψ_{j} the estimate for the norm is almost the best possible one for small β. Indeed, consider T = {1, 2} with the measure μ defined by

μ\{1\} = μ\{2\} = \frac{1}{2}. Then L^{p}(T) is merely ℝ^{2} with the norm ||(x_{1}, x_{2})||_{p} = (\frac{|x_{1}|^{p}+|x_{2}|^{p}}{2} )^{\frac{1}{p}}.

Let n be a large positive integer. Put ψ_{j} =^{def} \sqrt{\frac{2}{n}}(cos \frac{πj}{n} , sin \frac{πj}{n}) (j = 1, …, n). Note that

||ψ_{j}||_{2} = \sqrt{\frac{1}{n}} and therefore

∥∑_{j} c_{j}ψ_{j}∥_{2} ≤ \sqrt{\frac{1}{n}} ∑_{j} |c_{j}| ≤ (∑ c_{j}^{2})^{\frac{1}{2}} .

On the other hand, for every 1 ≤ q ≤ 2

||ψ_{j}||_{q} ≥ \sqrt{\frac{1}{2n}} =^{def} β.

Let now a_{j} = \sqrt{\frac{1}{n}} (j = 1, …, n). Notice that for every j = 1, …, n − 1 we have

||ψ_{j} − ψ_{j+1}||_{2} ≤ \sqrt{\frac{1}{n}} \frac{π}{n} and thereby for every p ≥ 2 we get |∫_{T} F ψ_{j}dμ − ∫_{T} F ψ_{j+1}dμ| ≤

\sqrt{\frac{1}{n}} \frac{π}{n}||F ||_{p} If the signs of ∫_{T} F ψ_{j}dμ and ∫_{T} F ψ_{j+1}dμ are different for some j, then, due to

the previous inequality, at least one of them doesn’t exceed \sqrt{\frac{1}{n}} \frac{π}{2n}||F ||_{p} and therefore

||F ||_{p} ≥ \frac{2n}{π} = \frac{1}{π} β^{−2}.

If all the signs are the same, we get the same result from ||ψ_{1} + ψ_{n}||_{2} ≤ \sqrt{\frac{1}{n}} \frac{π}{n}.

2) Another curious observation is that all the conditions (B_{p}) are equivalent if we do not care about the exact value of β. So, if any of the spaces L^{p}(T) is large, then all of them are !

Proof of the theorem:

Consider all sign corteges ε = {ε_{j}} where ε_{j} = +1 or −1. For each of them put f_{ε} =^{def} ∑_{j} ε_{j}a_{j}ψ_{j}. Regardless of the choice of the signs we have f_{ε} ∈ L^{2}(T) and ||f_{ε}||_{2} ≤ 1.

Let now Φ : ℝ → ℝ_{+} be any C^{2}-smooth even function satisfying Φ(0) = Φ′(0) = 0 and 0 < Φ′′(x) ≤ 1 for every x ∈ ℝ.

It is easy to check that the integral I(f) =^{def} ∫_{T} Φ(f)dμ is well-defined and continuous in L^{2}(T). As the family {f_{ε}} is compact in the topology of L^{2}(T), one can find a cortege ε̄ for which I(f_{ε}) attains its maximal value. Let f = fε̄. Fix now some j and consider the function f_{j} which is obtained from fε̄ by replacing ε̄_{j}a_{j}ψ_{j} by −ε̄_{j}a_{j}ψ_{j} in the corresponding sum. So f_{j} = f − 2ε̄_{j}a_{j}ψ_{j}.

We have

0 ≤ ∫_{T} Φ(f)dμ − ∫_{T} Φ(f_{j})dμ = ∫_{T} Φ′(f)(f − f_{j})dμ + \frac{1}{2} ∫_{T} Φ′′(g)(f − f_{j})^{2}dμ.

where g(t) lies between f(t) and f_{j}(t) for every t ∈ T. Recalling what f − f_{j} is, we get from here

ε̄_{j} ∫_{T} Φ′(f)ψ_{j}dμ ≥ a_{j} ∫_{T} Φ′′(g)ψ_{j}^{2}dμ.

If we contrive to choose Φ(x) in such a way that Φ′(f) ∈ L^{p}(T) for every f ∈ L^{2}(T) and that the integral on the right can be estimated from below by some constant depending on β only, it will remain only to put F =^{def} AΦ′(f) with constant A large enough.

\underline{Case p = 2.}

Here the choice is easy: just put Φ(x) = \frac{x^{2}}{2} . Then Φ′(f) = f and Φ′′(g) ≡ 1, so the

integral reduces to ∫_{T} ψ_{j}^{2} ≥ β^{2}. Thus, the function F =^{def} β^{−2}f is what we are looking for.

General case.

Let Φ(x) be the solution of the equation Φ′′(x) = (1 + x^{2})^{\frac{2}{p} −1} for which Φ(0) = Φ′(0) = 0. As p ≥ 2, we really have Φ′′(x) ≤ 1 for every x ∈ ℝ.

Note that

|Φ′(x)| ≤ ∫_{0}^{|x|}(1 + s^{2})^{\frac{2}{p} −1}ds ≤ (∫_{0}^{|x|} ds)^{\frac{2}{p}} (∫_{0}^{|x|}(1 + s^{2})^{−1}ds)^{1− \frac{2}{p}} ≤ (\frac{π}{2} )^{1− \frac{2}{p}} |x|^{\frac{2}{p}} .

Thus,

||Φ′(f)||_{p} ≤ (\frac{π}{2} )^{1− \frac{2}{p}} (∫_{T} |f|^{2}dμ)^{\frac{2}{p}} ≤ (\frac{π}{2} )^{1− \frac{2}{p}} .

On the other hand,

(∫_{T} (1 + g^{2})dμ)^{1− \frac{q}{2}} (∫_{T} (1 + g^{2})^{\frac{2}{p} −1}ψ_{j}^{2}dμ)^{\frac{q}{2}} ≥ ∫_{T} |ψ_{j}|^{q}

for 1 − \frac{q}{2} + (\frac{2}{p} − 1)\frac{q}{2} = 0.

As g^{2} ≤ f^{2} + f_{j}^{2}, we get ∫_{T} (1 + g^{2})dμ ≤ 3 and therefore

∫_{T} Φ′′(g)ψ_{j}^{2}dμ = ∫_{T} (1 + g^{2})^{\frac{2}{p} −1}ψ_{j}^{2}dμ ≥ 3^{1− \frac{2}{q}} (∫_{T} |ψ_{j}|^{q})^{\frac{2}{q}} ≥ 3^{1− \frac{2}{q}} β^{−2} = 3^{\frac{2}{p} −1}β^{−2}.

So, F = 3^{1− \frac{2}{p}} β^{−2}Φ′(f) is what we need.

THE BANG SOLUTION OF THE COEFFICIENT PROBLEM – 3

Beyond L^{∞} (written in Banach space geometry language)

Our main aim here will be to emphasize the geometric charachter of the problem and of our approach to it. Every time when we can make the geometry clearer by some loss in constants, we will do it. So, when doing L^{∞}-case again, we will get the estimate ≈ β^{−3} for the norm instead of the best possible β^{−2}. The reader may think himself over what should be changed to gain exactly the same result as before.

Let H be a Hilbert space over ℝ, B be a closed convex set, containing the origin (from this point we will call such sets “standard”). Denote by P_{B}f the nearest to f (in the metrics of H) element of B. We get a projection P_{B} : H → B (unfortunately not linear in general).

\underline{Lemma 1.}

For every f′, f′′ ∈ H we have ||P_{B}f′ − P_{B}f′′|| ≤ ||f′ − f′′||.

\underline{Proof:}

As B is convex, (f′ − P_{B}f′, g − P_{B}f′) ≤ 0 for every g ∈ B. In particular, (f′ − P_{B}f′, P_{B}f′′ − P_{B}f′) ≤ 0. Analogously, (f′′ − P_{B}f′′, P_{B}f′ − P_{B}f′′) ≤ 0. Thus,

(f′ − f′′, P_{B}f′ − P_{B}f′′)
= ||P_{B}f′ − P_{B}f′′||^{2} − (f′ − P_{B}f′, P_{B}f′′ − P_{B}f′) − (f′′ − P_{B}f′′, P_{B}f′ − P_{B}f′′)
≥ ||P_{B}f′ − P_{B}f′′||^{2}.

The statement of Lemma 1 follows from here immediately.

Actually, this all means just the following: if you have four points X, Y, Z, T in space and the angles XY Z and Y ZT are not less than \frac{π}{2} , then |Y Z| ≤ |XT |.

Thus, P_{B}f depends continuously on f (actually we proved that P_{B} ∈ Lip1). Let’s now show that it also depends continuously on B.

\underline{Lemma 2.}

Let B_{H} = {x ∈ H : ||x|| ≤ 1} be the unit ball in H. Let δ > 0, f ∈ H. If two standard sets B and B′ satisfy P_{B′}f ∈ B + δB_{H} and P_{B}f ∈ B′ + δB_{H} (i.e. P_{B′}f is not more than δ distant from B and vice versa), then

||P_{B}f − P_{B′}f|| ≤ \sqrt{2||f||δ}

\underline{Proof}

Let a = ||f − P_{B}f||, b = ||f − P_{B′}f||. Suppose, for definiteness, that a ≥ b. If a = 0, then P_{B}f = P_{B′}f = f and there is nothing to prove. Otherwise f and B lie in different half-spaces separated by the hyperplane orthogonal to f − P_{B}f and containing the point P_{B}f. Let H_{+} be the half-space containing f and H_{−} be the half-space, containing B. As a ≥ b, we have P_{B′}f ∈ H_{+}. Let u = ||P_{B}f − P_{B′}f||, v be the distance from P_{B′}f to H_{−}. The Pithagorean theorem gives v^{2} − (a − v)^{2} = u^{2} − b^{2}, or, what is the same, u^{2} = b^{2}−a^{2}+2av. As v doesn’t exceed the distance from P_{B′}f to B, which is not more than δ, and as b ≤ a, we get u ≤ \sqrt{2aδ}. Since 0 ∈ B, we have a = ||f − P_{B}f|| ≤ ||f − 0|| = ||f|| and the conclusion of Lemma 2 follows immediately.

The Bang functional Φ_{B}.

Let B be a standard set in H. Consider the functional Φ_{B} : H → ℝ defined by

Φ_{B}(f) = ||f||^{2} − ||f − P_{B}f||^{2} = 2(f, P_{B}f) − ||P_{B}f||^{2} (f ∈ H).

The functional Φ_{B} has the following remarkable

\underline{Property:}

Φ(f) − Φ(f′) ≤ 2(P_{B}f, f − f′) − ||P_{B}f − P_{B}f′||^{2}.

\underline{Proof:}

Φ(f) − Φ(f′) = 2(f, P_{B}f) − 2(f′, P_{B}f′) − ||P_{B}f||^{2} + ||P_{B}f′||^{2}
= 2(P_{B}f, f − f′) + 2(f′, P_{B}f − P_{B}f′) − (P_{B}f + P_{B}f′, P_{B}f − P_{B}f′)
= 2(P_{B}f, f − f′) + (2f′ − P_{B}f − P_{B}f′, P_{B}f − P_{B}f′)
= 2(P_{B}f, f − f′) + 2(f′ − P_{B}f′, P_{B}f − P_{B}f′) − ||P_{B}f − P_{B}f′||^{2}.

It remains only to note that the second term is nonpositive.

Observation:

Let {ψ_{j}} be a system of vectors in H, B be a standard set and, at last, {a_{j}} be a sequence of positive numbers.

Suppose that every cortege ε = {ε_{j}} where ε_{j} = 1 or −1 the series ∑_{j} ε_{j}a_{j}ψ_{j} converges in H to some element f_{ε}.

Consider the cortege ε̄ for which Φ_{B}(f_{ε}) attains its maximal value (as Φ_{B} is continuous and as convergence of all the series ∑_{j} ε_{j}a_{j}ψ_{j} implies their uniform convergence in H with respect to the choice of signs ε_{j}, the cortege ε̄ exists and can be constructed by diagonal process, say). Let f = fε̄ be the corresponding element of H. Fix some j consider the series in which ε̄_{j} is replaced by −ε̄_{j}. Let f′_{j} = f − 2ε̄_{j}a_{j}ψ_{j} be the sum of this series. Then

|(P_{B}f, ψ_{j})| ≥ \frac{||P_{B}f − P_{B}f′_{j}||^{2}}{4a_{j}}

\underline{Proof:}

We have 0 ≤ Φ_{B}(f) − Φ_{B}(f′_{j}) ≤ 2(P_{B}f, f − f′_{j}) − ||P_{B}f − P_{B}f′_{j}||^{2} which results in 2(P_{B}f, f − f′_{j}) ≥ ||P_{B}f − P_{B}f′_{j}||^{2} and, after recalling that f − f′_{j} = 2ε̄_{j}a_{j}ψ_{j}, in

ε̄_{j}(P_{B}f, ψ_{j}) ≥ \frac{||P_{B}f − P_{B}f′_{j}||^{2}}{4a_{j}} .

It remains to note that |ε̄_{j}| = 1.

Let now X = L^{∞}(T), H = L^{2}(T) where T is some measure space with probability measure μ. Let ψ_{j} be a system of vectors in H satisfying || ∑_{j} c_{j}ψ_{j}|| ≤ \sqrt{∑_{j} c_{j}^{2}} for every

cortege of coefficients c_{j} ∈ ℝ and obvious necessary condition ∫_{T} |ψ_{j}|dμ ≥ β > 0. At last, let a_{j} > 0 satisfy ∑_{j} a_{j}^{2} = 1.

Then for every cortege ε we have the corresponding series convergent in L^{2}(T) to the sum f_{ε} of norm ||f_{ε}||_{H} ≤ 1. Let s > 0, B = sB_{L^{∞}(T)}. The projection P_{B}f is easy to compute for every f ∈ H. Namely,

(P_{B}f)(t) = \{ \begin{matrix}−s, & if f(t) ≤ −s; \\ f(t), & if − s ≤ f(t) ≤ s; \\ s, & if f(t) ≥ s \end{matrix} (t ∈ T).

Thus, (for the same f and f′ as above), ||P_{B}f − P_{B}f′||^{2} ≥ ∫_{E} |f(t) − f′(t)|^{2}dμ(t) = 4a_{j}^{2} ∫_{E} ψ_{j}^{2}dμ, where E = {t ∈ T : |f(t)| ≤ s, |f′(t)| ≤ s}. The measure of the complement T \ E of the set E can be estimated from above by \frac{2}{s^{2}} (just from Tschebyshev inequality). On the other hand, if μ(T \ E) ≤ \frac{β^{2}}{4} , then

∫_{T \E} |ψ_{j}|dμ ≤ μ(T \ E)^{\frac{1}{2}} (∫_{T} ψ_{j}^{2}dμ)^{\frac{1}{2}} ≤ \frac{β}{2}

and thereby

∫_{E} ψ_{j}^{2}dμ ≥ (∫_{E} |ψ_{j}|dμ)^{2} ≥ \frac{β^{2}}{4}

Thus, if we put s = \frac{4}{β}, say, we will have

|(P_{B}f, ψ_{j})| ≥ \frac{β^{2}}{4} a_{j}

for every j. Then the function F = \frac{4}{β^{2}} P_{B}f has L^{∞}-norm ||F ||_{L^{∞}(T)} ≤ \frac{16}{β^{3}} and its Fourier coefficients with respect to ψ_{j} are greater than a_{j} in absolute value. So, L^{∞}(T) is large in L^{2}(T).

From boundedness to continuity.

If T is a topological space and the space C(T) of bounded continuous functions (with the same norm as in L^{∞}) is dense in L^{2}(T), there is a temptation to improve the constructed function g with large Fourier coefficients to a continuous one.

Indeed, if the set {ψ_{j}} is finite, all we need is to take the new function F′ ∈ C(T) sufficiently close to F in L^{2}-norm. We can do it for any finite system with uniform estimate for the norm, the space C(T) is complete and therefore … nothing follows ! One of many strange things connected with the problem is that nobody can prove or disprove the result that general (namely, it is not known whether it is always enough to check that X is large with respect to all finite subsystems of the system {ψ_{j}} (which you may assume even exactly orthogonal) with uniform estimate for the norm to conclude that X is large).

So, instead of referring to some general principle, we have to work by hand. Let us consider again the standard set B = sB_{L^{∞}(T)} with s = \frac{4}{β} as before, the sequence of numbers a_{j} > 0 satisfying ∑_{j} a_{j}^{2} = 1 and the corresponding family {f_{ε} = ∑_{j} ε_{j}a_{j}ψ_{j} : ε_{j} = 1 or − 1}. Note that the family {f_{ε}} is compact and thereby the set {P_{B}f_{ε}} is compact as well (both in topology of L^{2}(T), of course). As sB_{C(T)} is dense in B = sB_{L^{∞}(T)} (again, in L^{2}-sense), we can find for every δ > 0 a finite set of elements g_{1}, …, g_{N(δ)} ∈ sB_{C(T)} such that for every cortege ε the element P_{B}f_{ε} is not more than δ distant from one of g_{m}. Let B(δ) be the convex hull of the origin and the points g_{m}. Then B(δ) is a standard set for every δ > 0 and B(δ) ⊂ sB_{C(T)} ⊂ B (the main advantage of B(δ) compared to sB_{C(T)} is that B(δ) is closed in topology of L^{2}(T)). Besides, every element P_{B}f_{ε} is not more than δ distant from B(δ). Fix some ε and consider again the pair of functions f = f_{ε} and f′_{j} = f − 2ε_{j}a_{j}ψ_{j}. We have

||P_{B(δ)}f − P_{B(δ)}f′_{j}|| ≥ ||P_{B}f − P_{B}f′_{j}|| − ||P_{B(δ)}f − P_{B}f|| − ||P_{B(δ)}f′_{j} − P_{B}f′_{j}||.

But, as we saw above, the first term on the right is at least βa_{j} while lemma 2 together with the estimates ||f||, ||f′_{j}|| ≤ 1 implies that both other tems do not exceed \sqrt{2δ}. Thus,

if δ ≤ \frac{β^{2}a_{j}^{2}}{32} , we still have

||P_{B(δ)}f − P_{B(δ)}f′_{j}|| ≥ \frac{β}{2} a_{j}.

Let us now consider some sequence δ_{k} → 0 and define the functional Φ : H → ℝ to maximize by

Φ(f) = ∑_{k=1}^{∞} \frac{1}{k(k + 1)}Φ_{B(δ_{k})}(f).

Again, let ε̄ be the cortege for which Φ attains its maximal value. For the functions f = fε̄ and f′_{j} = f − 2ε̄_{j}a_{j}ψ_{j} we get

0 ≤ Φ(f) − Φ(f′_{j}) ≤ ∑_{k=1}^{∞} \frac{1}{k(k + 1)}[2(P_{B(δ_{k})}f, f − f′_{j}) − ||P_{B(δ)}f − P_{B(δ)}f′_{j}||^{2}]

Thus, for the function F = ∑_{k} \frac{1}{k(k+1)}P_{B(δ)}f (which clearly is in C(T)) we have

ε̄_{j}(F, ψ_{j}) ≥ \frac{1}{4a_{j}} ∑_{k=1}^{∞} \frac{1}{k(k + 1)}||P_{B(δ)}f − P_{B(δ)}f′_{j}||^{2}.

But

∑_{k=1}^{∞} \frac{1}{k(k + 1)}||P_{B(δ)}f − P_{B(δ)}f′_{j}||^{2} ≥ ∑ ; k : δ_{k} ≤ \frac{β^{2}a_{j}^{2}}{32} \frac{1}{k(k + 1)} \frac{β^{2}}{4} a_{j}^{2} = \frac{β^{2}a_{j}^{2}}{4k(a_{j})},

where for every a > 0 the index k(a) is defined as the first index k for which δ_{k} ≤ \frac{β^{2}a_{j}^{2}}{32} . This results in

|(F, ψ_{j})| ≥ \frac{β^{2}}{16} \frac{a_{j}}{k(a_{j})}

for every j. As the sequence δ_{k} can tend to 0 as fast as we desire, k(a) is just an arbitrary decreasing finction which has values 1, 2, 3, … and tends to infinity as a → 0. Therefore the sequence \{\frac{a_{j}}{k(a_{j})}\} is nothing but an arbitrary sequence b_{j} of positive numbers satisfying ∑_{j} b_{j}^{2} < 1. Thus, C(T) is large and we have done.

How about U(𝕋) ?

For a while I thought that I knew how to get the result that the space U(𝕋) (i.e. the space of functions defined on the unit circumference 𝕋 for which their Fourier series with respect to the standard base z^{j} converges uniformly with norm defined as the supremum of L^{∞}-norms of partial sums) is large under the same assumptions about ψ_{j} as before, but then I discovered that I made quite rough (though very hard to notice) mistake in my reasoning. Nevertheless, the “simple” part of what I was doing has survived and gives the following result which is only a little weaker:

Proposition:

Suppose that a sequence of numbers a_{j} ∈ (0, \frac{1}{2}), say, satisfies the condition

∑_{j} a_{j}^{2}| log a_{j}|^{γ} < ∞ with some γ > 2.

Then there is a function F ∈ U(𝕋) for which |(F, ψ_{j})| ≥ a_{j} for every j.

\underline{Proof:}

Let us do the same as for the case X = C(T). The main difference is that now we can approximate an element P_{B}f_{ε} ∈ B = sB_{L^{∞}(𝕋)} by an element g ∈ U(𝕋) only if we allow the norm ||g||_{U(𝕋)} to grow to infinity. Fortunately, we do not need it to grow very fast. The best possible estimate is given by the famous

Kislyakov correction theorem:

If ||f||_{L^{∞}(𝕋)} ≤ 1, then for every δ ∈ (0, \frac{1}{2}), say, there exists g ∈ U(𝕋) such that ||f − g||_{L^{2}(𝕋)} ≤ δ while ||g||_{U(𝕋)} ≤ C log \frac{1}{δ} (C > 0 is some absolute constant).

We can now repeat the construction of standard sets B(δ) using the Kislyakov theorem instead of trivial approximation of bounded functions by continuous ones. For every δ > 0 we see that B(δ) ⊂ B + δB_{L^{2}(𝕋)} and the U(𝕋)-norms of all elements in B(δ) do not exceed Cs log \frac{s}{δ}. It allows to repeat all the estimates except that for the norm of the element F constructed. Now, to get the series ∑_{k} \frac{1}{k(k+1)}P_{B(δ)}f convergent in U(𝕋) we have to demand ∑_{k} \frac{1}{k(k+1)} log \frac{1}{δ_{k}} < ∞.

We will just put δ_{k} = exp\{−k^{\frac{2}{γ}} \} to provide that (any further advance which we could gain from a better choice is hardly noticable compared to the distance to the desired result that U(𝕋) is large). This gives k(a) ≈ | log a|^{\frac{γ}{2}} for small a and we conclude that for every sequence of numbers a_{j} ∈ (0, \frac{1}{2}) satisfying ∑_{j} a_{j}^{2} = 1 there is a function F ∈ U(𝕋) such that for every j

|(F, ψ_{j})| ≥ \frac{a_{j}}{| log a_{j}|^{\frac{γ}{2}}} ,

which is equivalent to the Proposition to prove.

THE BANG SOLUTION OF THE COEFFICIENT PROBLEM – 4 How Bang solved the plank problem. Let B ⊂ ℝ^{n} be a compact convex body with smooth boundary covered by finitely many

open strips S_{j} (general case can be easily reduced to this one by standard approximation argument). Let H be the width of B, h_{j} be the width of S_{j}. Put a_{j} =^{def} \frac{h_{j}}{2} and consider the vectors ψ_{j} of length a_{j} orthogonal to the boundary hyperplanes of the corresponding strips S_{j}.

Bang’s solution consists of two independent steps:

\underline{Lemma 1:}

If ψ is a vector of length a < \frac{H}{2} , then the intersection (B − ψ) ∩ (B + ψ) contains a homotetic image of B with coefficient \frac{H−2a}{H}

\underline{Corollary:}

If ∑_{j} h_{j} < H, then there exists a point x_{0} ∈ ℝ^{n} such that for every sign cortege

ε = {ε_{j}} the point x_{ε} =^{def} x_{0} + ∑_{j} ε_{j}ψ_{j} ∈ B.

\underline{Lemma2:}

For every x_{0} ∈ ℝ^{n} at least one of the points x_{ε} doesn’t belong to any of the strips S_{j}.

Proof of lemma 1:

Let yz be the longest section of B by a line parallel to ψ. Let ℓ be the length of the interval yz, c be the middle of this interval. Note that the tangent hyperplanes to B at the points y and z are parallel (othewise we might move the line yz a bit and get a longer section). Since ℓ is not less than the distance between these hyperplanes, we conclude that ℓ ≥ H.

As ψ = \frac{a}{ℓ}(z − y), for every x ∈ B we have

\frac{2a}{ℓ}c + \frac{ℓ − 2a}{ℓ}x + ψ = \frac{2a}{ℓ} \frac{y + z}{2} + \frac{ℓ − 2a}{ℓ}x + \frac{a}{ℓ}(z − y) = \frac{2a}{ℓ}z + \frac{ℓ − 2a}{ℓ}x ∈ B

(because B is convex, x, z ∈ B and \frac{2a}{ℓ} + \frac{ℓ−2a}{ℓ} = 1). Therefore, \frac{2a}{ℓ}c + \frac{ℓ−2a}{ℓ}x ∈ B − ψ. Analogously we get that also \frac{2a}{ℓ}c + \frac{ℓ−2a}{ℓ}x ∈ B + ψ. It remains to note that the point \frac{2a}{ℓ}c + \frac{ℓ−2a}{ℓ}x runs over the image of B under the homotopy with the center c and the coefficient \frac{ℓ−2a}{ℓ} ≥ \frac{H−2a}{H} as x runs over B.

Proof of the corollary:

By induction in the number of vectors ψ_{j} we find that the intersection ∩_{ε}(B − ∑_{j} ε_{j}ψ_{j})

contains a honotopic image of B with the coefficient ^{\frac{H−∑_{j} h_{j}}{H}} > 0 and therefore is not empty. Every point of this intersection can be taken as x_{0}.

Remark to the step 1:

We didn’t use this part of Bang’s solution in our reasoning at all because in our case we had the body B symmetric with respect to the origin which allows just to put x_{0} = 0. Also, it is this part where the distinction between ∑_{j} a_{j} in the plank problem and ∑_{j} a_{j}^{2} in the coefficient problem comes from: in the plank problem we deal with completely arbitrary vectors ψ_{j} and therefore the triangle inequality is the only estimate for the norm we can use, while in the coefficient problem we had our vecrors mutually orthogonal and might apply the Pithagorean theorem instead.

Proof of lemma 2:

Let p_{k} be any point of the “middle” hyperplane of the strip S_{k}. Then

S_{k} = {x ∈ ℝ^{n} : |(x − p_{k}, ψ_{k})| < a_{j}^{2}.

As adding ψ_{j} to a point means pushing the point toward the “positive” side of the strip S_{j} while adding −ψ_{j} means pushing it in the opposite direction, it is natural to try to find the point x_{ε} which lies from the positive or from the negative side of S_{k} corresponding to whether ε_{k} = 1 or −1. This gives the system

ε_{k}(x_{ε} − p_{k}, ψ_{k}) ≥ a_{k}^{2} (k = 1, 2, …)

of inequalities to solve. Recalling that x_{ε} = x_{0} + ∑_{j} ε_{j}ψ_{j}, we get

ε_{k}(x_{0} − p_{k}, ψ_{k}) + ∑_{j} ε_{k}ε_{j}(ψ_{k}, ψ_{j}) ≥ a_{k}^{2} (k = 1, 2, …).

Note that when j = k, we have the term ε_{k}^{2}|ψ_{k}|^{2} = a_{k}^{2} on the left. Cancelling it out, we get left with

ε_{k}(x_{0} − p_{k}, ψ_{k}) + ∑_{j:j≠k} ε_{k}ε_{j}(ψ_{k}, ψ_{j}) ≥ 0 (k = 1, 2, …).

Let Φ(ε) =^{def} ∑_{k}ε_{k}(x_{0} − p_{k}, ψ_{k}) + ∑_{j,k:j<k} ε_{k}ε_{j}(ψ_{k}, ψ_{j}). Note that the inequality above is

exactly \frac{1}{2}(Φ(ε) − Φ(ε′)) ≥ 0 where ε′_{j} = ε_{j} for every j ≠ k and ε′_{k} = −ε_{k}. Thus, if we take the sign cortege ε̄ for which the function Φ(ε) attains its maximal value, the corresponding point xε̄ will lye in none of the strips. That’s all !

THE BANG SOLUTION OF THE COEFFICIENT PROBLEM – 5 Concluding remarks (written a year later). This paper was written about a year ago. I intentionally delayed with its publication

hoping that someone would be able to combine Bang’s approach with some other idea (say, with the classical Kahane–Katznelson–de-Leeuw technique) and to get the result that U(𝕋) is large with respect to a system {ψ_{j}} satisfying the same a priori assumption ∥ ∑_{j} a_{j}ψ_{j}∥_{2} ≤ (∑_{j} a_{j}^{2})^{1/2} and the trivial necessary condition ∫_{𝕋} |ψ_{j}|dμ ≥ β > 0. Despite the paper has got quite famous during that year and despite all my own efforts, this has not happened yet. Moreover, even a simpler question about “largeness” of X = H^{∞}(𝔻) in H = H^{2}(𝔻) remains unanswered.

Still I’ve got several interesting remarks from various people. Two most interesting of them were the following.

First, Keith Ball pointed out to me that the “coefficient problem” can be solved completely if one is interested in majorizing a ℓ^{1} sequence {a_{j}} instead of ℓ^{2} that. Namely, in 1991 he proved the following

\underline{Theorem.}

Let X be any Banach space, ψ_{j} ∈ X^{∗} be any family of bounded linear functionals on X such that ∥ψ_{j}∥_{X^{∗}} = 1 for every j. Then for any sequence {a_{j}} of positive numbers satisfying ∑_{j} a_{j} < 1 there exists f ∈ X such that ∥f∥_{X} ≤ 1 and |(f, ψ_{j})| ≥ a_{j} for every j.

The proof of the theorem can be found in [3].

The second remark is due to Francoise Lust-Piquard. She noticed that the Bang approach works also in a “non-commutative” setting. Among other results her forthcoming paper [4] contains the following

Proposition.

Let {a_{ij}}_{i,j=1}^{∞} be a matrix with positive entries such that

∑_{j=1}^{∞} a_{ij}^{2} ≤ 1 for every i and ∑_{i=1}^{∞} a_{ij}^{2} ≤ 1 for every j.

Then there exists a bounded operator A : ℓ^{2} → ℓ^{2} of norm ∥A∥ ≤ 10, say, such that its matrix satisfies |A_{ij}| ≥ a_{ij} for every i, j.

The above condition is obviously necessary for existence of such an operator of norm 1.

Literature cited

[1] Bang T., A solution of the ”plank problem.” Proc. Amer. Math. Soc. 2, (1951). 990–993

[2] de Leeuw, K; Katznelson, Y; Kahane, J.-P. Sur les coefficients de Fourier des fonctions continues. C. R. Acad. Sci. Paris Sr. A-B 285 (1977), no. 16, A1001–A1003

[3] Ball K. The plank problem for symmetric bodies. Invent. Math. 104 (1991), no. 3, 535–543

[4] Lust-Piquard, F. On the coefficient problem: a version of the Kahane-Katznelson-De Leeuw theorem for spaces of matrices. to appear in Journal of Functional Analysis.