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\centerline{\bf THE BANG SOLUTION OF THE COEFFICIENT PROBLEM -- 1}
\centerline{\bf Possible introduction (written in common English).}
\medskip
\rightline{\it They call themselves Analysts, but what they do}
\rightline{\it is just manipulating inequalities !...}
\rightline{(ascribed to Arnold)}
\bigskip
Let $T$ be a measure space with probability measure $\mu$.
Let $\psi_j:T\to\R$ be an orthonormal system of functions in $L^2(T)$,
i.e. $||\psi_j||_2=1$ for every $j$ and $(\psi_i,\psi_j)\ed\int_T\psi_i
\psi_jd\mu=0$ for every $i\ne j$. Suppose that $X$ is some other space
of functions on $T$ such that $||f||_2\le||f||_X$ for every $f\in X$
(the main example to keep in mind is $X=L^\infty(T)$).
The question we are going to discuss is whether we can say anything
nontrivial about the decay of the Fourier coefficients
$(f,\psi_j)=\int_Tf\psi_jd\mu$ of functions $f\in X$.
Of course, we always have the Bessel inequality:
$$
\sum_j(f,\psi_j)^2\le||f||_2^2\le||f||_X^2.
$$
\ul{Definition }
{\it We will call the space $X$ large
(with respect to the system $\{\psi_j\}$)
if for every sequence $\{a_j\}$ of positive
numbers satisfying $\sum_ja_j^2=1$ there exists a function $F\in X$ such
that $|(F,\psi_j)|\ge a_j$ for every $j$.}
\medskip
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 $|(F,\psi_j)|\ge a_j$ with
uniformly bounded norm (i.e. such that $||F||_X\le C$ where $C>0$ does
not depend on $a_j$). Indeed, otherwise for every $k=1,2,\dots$ one
can choose a sequence $\{a_j^{(k)}\}$ with $\sum_j(a_j^{(k)})^2=1$
and such that there is no function $F\in X$ with $||F||_X\le 4^k$ satisfying
$(F,\psi_j)|\ge a_j^{(k)}$ for all $j$. But then for the sequence
$a_j\ed\sum_{k=0}^\infty2^{-k}a_j^{(k)}$ there is no $F\in X$ solving
the coefficient problem $|(F,\psi_j)|\ge a_j$ (because $a_j\ge 2^{-k}
a_j^{(k)}$ for every $k$ and thereby we should have $||F||_X\ge 2^k$
for every $k$ which is impossible), but still $\sum_ja_j^2\le1$.
Which spaces are large and which are not ? There exists the following
\ul{Conjecture:}
{\it $X$ is large iff there is a constant $c>0$ such that for every sequence
$\{b_j\}$ of positive numbers satisfying $\sum_jb_j^2=1$ one can find
$f\in X$ such that
$$
||f||_X=1 \text{\ \ and \ }\sum_jb_j|(f,\psi_j)|\ge c
.$$}
\medskip
The conjecture means that the set of sequences $\{\,\{|(f,\psi_j)|\}:
f\in X, ||f||_X\le 1\}$ (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 $\{z^k\}_{k=-\infty}^\infty$ (the norm in $U(\T)$
is defined by $||f||_U(\T)\ed
\sup_{n,m\in\Bbb Z}||S_{n,m}f(z)||_\infty$ where $S_{n,m}f(z)\ed
\sum_{k=n}^m\hat f(k)z^k$). It is known that $U(T)$ is large with
respect to any orthonormal system $\{\psi_j\}$ satisfying $\sup_j
||\psi_j||_\infty<+\infty$, 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 $f_\e\ed\sum_j\e_ja_j\psi_j$ (where $\e_j=1$ or $\e_j=-1$
with probability $\frac12$ 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 $f_\e$ into the
sum $f_\e=g_A+h_A$ such that $||g_A||_X\le A$, $||h_A||_2\le\D(A)$ and
$\D(A)$ decays fast enough as $A\to+\infty$.
Then they proceeded by ``minor corrections'' as follows:
Suppose that we have a function $F\in X$ of norm $||F||_X\le M$
which solves the coefficient problem for all $j$ but those in some
exceptional set $J$ and that $\sum_{j\in J}a_j^2\le b^2$ (for instance, we may
start with $F\equiv 0$, $M=0$, $J$ which is the whole index set and $b=1$).
Let us choose the signs $\e_j$ in such a way that $f_\e=\sum_{j\in J}\e_j
a_j\psi_j$ almost belongs to $X$. Note that the function $\widetilde F\ed F+2f_\e$
solves the coefficient problem $|(\widetilde F,\psi_j)|\ge a_j$ 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 $f_\e$ into the sum
$g_A+h_A$ where $||g_A||_X\le Ab$ and $||h_A||_2\le \D(A)b$ (the factor $b$ is
due to the condition $\sum_{j\in J}a_j^2\le b^2$, not $1$, as in the
definition of $\D(A)$ above), the functuon $\widetilde F$ can be decomposed into the
sum of $F_A\ed F+g_A$ and $h_A$. When we take $F_A$ instead of $\widetilde F$,
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
$c\in (0,1)$ and let $J'=\{j:|(F_A,\psi_j)|<(1-c)a_j\}$. Then for every $j\in
J'$ we have $|(h_A,\psi_j)|\ge ca_j$ and therefore
$$
\sum_{j\in J'}a_j^2\le c^{-2}||h_A||_2^2\le\frac{\D(A)^2b^2}{c^2}\ed {b'}^2.
$$
For $j\notin J'$ the coefficients $(F_A,\psi_j)$ are easy to repair: just take
$F'\ed\frac{1}{1-c}F_A$. Thus, having started with the function $F\in X$ of
norm $||F||_X\le M$ and with the exceptional set $J$ for which
$\sum_{j\in J}a_j^2\le b^2$, we got a new function $F'$ of norm $||F'||_X\le
M'\ed \frac{M+Ab}{1-c}$ and a new exceptional set $J'$ for which
$\sum_{j\in J'}a_j^2\le{b'}^2$ where $b'=\frac{\D(A)b}{c}$. If we contrive to
get $b'**0$ 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 $\sum_kb_kA(b_{k+1})<+\infty$.
The trivial estimate
$$
\sum_kb_kA(b_{k+1})>\sum_k(b_k-b_{k+1})A(b_{k+1})\ge\int_0^1A(t)dt
$$
shows that we should assume at least that $\int_0A(t)dt<+\infty$.
On the other hand, whatever $\D$ is, $A(t)$ turns out to be quite regular
function. Indeed, since $\frac t2(2A(t))=tA(t)=\D(A(t))\ge\D(A(\frac t2))$,
we have $A(\frac t2)\le 2A(t)$ for every $t>0$.
Thus, taking $b_k=2^{-k}$, we get
$$
\sum_{k=0}^\infty b_kA(b_{k+1})=\sum_{k=0}^\infty 2^{-k}A(2^{-k-1})\le
4\int_0^1A(t)dt.
$$
So, the condition $\int_0A(t)dt<+\infty$ is also sufficient.
What does it mean in terms of the original function $\D$ ? Changing
variable from $t$ to $A$, we get
$\int^\infty Ad\left(\frac{\D(A)}{A}\right)<+\infty$ and it remains only to
integrate by parts to obtain the celebratred
\ul{De-Leeuw -- Kahane -- Katznelson theorem:}
{\it If $\dsize{\int^\infty\frac{\D(A)}{A}dA<+\infty}$, then $X$ is large
in $L^2(T)$ with respect to the system $\psi_j$. }
For instance, when $X=L^\infty(T)$ and the norms
$||\psi_j||_\infty$ are uniformly bounded, we may use the Khinchin theorem
to get that $f_\e\in L^4$, say, with estimate for the norm $||f_\e||_4$
not depending on the sequence $\{a_j\}$. This gives $\D(A)\le CA^{-1}$
which is more than enough to apply the theorem and to conclude that
$L^\infty(T)$ 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
\smallskip
\noindent 1) {\it $X$ is dense in $L^2(T)$;}
\noindent 2) {\it All $\psi_j$ almost belong to $X$ themselves.}
\smallskip
Correspondently, there remained two problems which could not be solved by the
de-Leeuw -- Kahane -- Katznelson approach:
\ul {1. The "support" problem:}
{\it Let $T'$ be a subset of $T$ of positive measure. Is the space
$X=L^p(T')\ed\{f\in L^p(T): supp\,f\subset T'\}$ ($2\le p\le\infty$)
large or not~?}
and
\ul {2. The ``minimal assumptions'' problem:}
{\it Let again $X=L^p(T)$. What are the minimal requirements on
the system $\{\psi_j\}$ which garantee that $X$ is large with
respect to this system ?}
\medskip
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
$\psi_j=z^j$ ($j\in\Bbb Z$)! Of course, the density of $X$ in $L^2(T)$
can be restored if we completely forget about $T\setminus T'$ and consider
$T'$ as the whole space (with the measure $d\mu'\ed\frac{d\mu}{\mu(T')}$).
Then we should also forget about the the original system $\psi_j$ and
consider the functions $\psi'_j\ed\sqrt{\mu(T')}\psi_j|_E$.
(the renormalization which preserves $L^2$-norm) instead. Unfortunately the
system $\{\psi'_j\}$ is no longer orthogonal. In general it can be just an
arbitrary system of functions satisfying
$||\sum_jc_j\psi'_j||_{L^2(T',\mu')}\le\left(\sum_jc_j^2\right)^{\frac12}$.
And it is also fatal: first, now we cannot say that the coefficients
$(f_\e,\psi'_j)$ of the sum $f_\e=\sum_j\e_ja_j\psi_j$ are just $\e_ja_j$ and
second, we no longer have a possibility to correct one coefficient without
substantial influencing others (just because $\psi'_j$ 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 $\psi_j\in
L^2\log^{2+\d}L$, but there is no evident reason for a condition of this kind
to be necessary (and actually it is not).
\ul{What are we going to do ?}
Consider the sets $S_j\ed\{f\in L^2(T):|(f,\psi_j)|0$. For the case $X=L^p(T)$
this means
$$
\biggl(\int_T|\psi_j|^q\biggr)^{\frac1q}\ge\b>0
\text{\qquad for every }j\qquad (B_p)
$$
where $q$ is the exponent conjugate to $p$, i.e. $\frac1p+\frac1q=1$.
Our aim is to show that $(B_p)$ is also sufficient. More precisely, we
have the following
\ul{Theorem:}
If the system $\{\psi_j\}$ satisfies $(B_p)$, then for every sequence of
positive numbers $a_j$ satisfying $\sum_ja_j^2=1$ we can find a
function $F\in L^p(T)$, such that
$$
||F||_p\le\left(\frac{3\pi}2\right)^{1-\frac2p}\b^{-2}\le5\b^{-2}
\text{\quad and \quad}\left|\int_TF\psi_jd\mu\right|\ge a_j
\text{ for every }j.
$$
\ul{Remark:}
1) Without additional assumptions on $\psi_j$ the estimate for the
norm is almost the best possible one for small $\b$. Indeed, consider
$T=\{1,2\}$ with the measure $\mu$ defined by $\mu\{1\}=\mu\{2\}=\frac12$.
Then $L^p(T)$ is merely $\R^2$ with the norm
$||(x_1,x_2)||_p=\left(\frac{|x_1|^p+|x_2|^p}2\right)^{\frac1p}$.
Let $n$ be a large positive integer.
Put $\psi_j\ed\sqrt{\frac2n}(\cos\frac{\pi j}n, \sin\frac{\pi j}n)$
($j=1,\dots,n$). Note that $||\psi_j||_2=\sqrt{\frac1n}$ and therefore
$$
\biggl\Vert\sum_jc_j\psi_j\biggr\Vert_2\le\sqrt{\frac1n}
\sum_j|c_j|\le\left(\sum
c_j^2\right)^{\frac12}.
$$
On the other hand, for every $1\le q\le 2$
$$
||\psi_j||_q\ge\sqrt{\frac1{2n}}\ed\b.
$$
Let now $a_j=\sqrt{\frac1n}$ ($j=1,\dots,n$). Notice that for every
$j=1,\dots,n-1$ we have $||\psi_j-\psi_{j+1}||_2\le\sqrt{\frac1n}
\frac\pi n$ and thereby for every $p\ge2$ we get
$
\left|\int_TF\psi_jd\mu-\int_TF\psi_{j+1}d\mu\right|\le
\sqrt{\frac1n}
\frac\pi n||F||_p
$
If the signs of $ \int_TF\psi_jd\mu$ and $\int_TF\psi_{j+1}d\mu $ are different
for some $j$, then, due to the previous inequality, at least one of
them doesn't exceed $\sqrt{\frac1n}
\frac\pi {2n}||F||_p $ and therefore
$$
||F||_p\ge\frac{2n}\pi =\frac1{\pi}\b^{-2}.
$$
If all the signs are the same, we get the same result from $||\psi_1+\psi_n||_2
\le \sqrt{\frac1n}
\frac\pi n$.
\medskip
2) Another curious observation is that all the conditions $(B_p)$ are
equivalent if we do not care about the exact value of $\b$. So, if
any of the spaces $L^p(T)$ is large, then all of them are~!
\ul{Proof of the theorem:}
Consider all sign corteges $\e=\{\e_j\}$ where $\e_j=+1$ or $-1$.
For each of them put $f_\e\ed\sum_j\e_ja_j\psi_j$. Regardless of the
choice of the signs we have $f_\e\in L^2(T)$ and $||f_\e||_2\le 1$.
Let now $\Phi:\R\to\R_+$ be any $C^2$-smooth even function satisfying
$\Phi(0)=\Phi'(0)=0$ and $0<\Phi''(x)\le 1$ for every $x\in\R$.
It is easy to check that the integral $I(f)\ed\int_T\Phi(f)d\mu$ is
well-defined and continuous in $L^2(T)$. As the family $\{f_\e\}$
is compact in the topology of $L^2(T)$, one can find a cortege $\bar\e$
for which $I(f_\e)$ attains its maximal value. Let $f=f_{\bar\e}$.
Fix now some $j$ and consider the function $f_j$ which is obtained
from $f_{\bar\e}$ by replacing $\bar\e_ja_j\psi_j$ by $-\bar\e_ja_j\psi_j$
in the corresponding sum. So $f_j=f-2\bar\e_ja_j\psi_j$.
We have
$$
0\le\int_T\Phi(f)d\mu-\int_T\Phi(f_j)d\mu=
\int_T\Phi'(f)(f-f_j)d\mu+\frac12\int_T\Phi''(g)(f-f_j)^2d\mu.
$$
where $g(t)$ lies between $f(t)$ and $f_j(t)$ for every $t\in T$.
Recalling what $f-f_j$ is, we get from here
$$
\bar\e_j\int_T\Phi'(f)\psi_jd\mu\ge a_j\int_T\Phi''(g)\psi_j^2d\mu.
$$
If we contrive to choose $\Phi(x)$ in such a way that $\Phi'(f)\in L^p(T)$
for every $f\in L^2(T)$ and that the integral on the right can be
estimated from below by some constant depending on $\b$ only, it will
remain only to put $F\ed A\Phi'(f)$ with constant $A$ large enough.
\ul{Case $p=2$.}
Here the choice is easy: just put $\Phi(x)=\frac{x^2}2$. Then
$\Phi'(f)=f$ and $\Phi''(g)\equiv 1$, so the integral reduces to
$\int_T\psi_j^2\ge\b^2$. Thus, the function $F\ed\b^{-2}f$ is what
we are looking for.
\ul{General case.}
Let $\Phi(x)$ be the solution of the equation
$\Phi''(x)=(1+x^2)^{\frac2p-1}$ for which $\Phi(0)=\Phi'(0)=0$.
As $p\ge 2$, we really have $\Phi''(x)\le 1$ for every $x\in\R$.
Note that
$$
|\Phi'(x)|\le\int_0^{|x|}(1+s^2)^{\frac2p-1}ds\le
\left(\int_0^{|x|}ds\right)^{\frac2p}\left(\int_0^{|x|}(1+s^2)^{-1}ds\right)
^{1-\frac2p}\le\left(\frac\pi2\right)^{1-\frac2p}|x|^{\frac2p}.
$$
Thus,
$$
||\Phi'(f)||_p\le\left(\frac\pi2\right)^{1-\frac2p}
\left(\int_T|f|^2d\mu\right)^{\frac2p}\le\left(\frac\pi2\right)^{1-\frac2p}.
$$
On the other hand,
$$
\left(\int_T(1+g^2)d\mu\right)^{1-\frac q2}
\left(\int_T(1+g^2)^{\frac2p-1}\psi_j^2d\mu\right)^{\frac q2}\ge
\int_T|\psi_j|^q
$$
for $1-\frac q2+(\frac 2p-1)\frac q2=0$.
As $g^2\le f^2+f_j^2$, we get $\int_T(1+g^2)d\mu\le3$
and therefore
$$
\int_T\Phi''(g)\psi_j^2d\mu=\int_T(1+g^2)^{\frac2p-1}\psi_j^2d\mu
\ge3^{1-\frac2q}\left(\int_T|\psi_j|^q\right)^\frac2q\ge
3^{1-\frac2q}\b^{-2}=3^{\frac2p-1}\b^{-2}.
$$
So, $F=3^{1-\frac2p}\b^{-2}\Phi'(f)$ is what we need.
\bigskip
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\def\ul#1{\medskip\noindent\underbar{\bf #1}\medskip}
\centerline{\bf THE BANG SOLUTION OF THE COEFFICIENT PROBLEM -- 3}
\centerline{\bf Beyond $L^\infty$ (written in Banach space geometry language)}
\medskip
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 $L^\infty$-case again, we will get the estimate
$\approx\b^{-3}$ for the norm instead of the best possible $\b^{-2}$.
The reader may think himself over what should be changed to gain exactly
the same result as before.
\medskip
Let $H$ be a Hilbert space over $\Bbb R$, $B$ be a closed convex set,
containing the origin (from this point we will call such sets ``standard'').
Denote by $P_Bf$ the nearest to $f$ (in the metrics of $H$) element of $B$.
We get a projection $P_B:H\to B$ (unfortunately not linear in general).
\ul{Lemma 1.}
For every $f',f''\in H$ we have $||P_Bf'-P_Bf''||\le||f'-f''||$.
\ul{Proof:}
As $B$ is convex, $(f'-P_Bf',g-P_Bf')\le0$ for every $g\in B$.
In particular, $(f'-P_Bf',P_Bf''-P_Bf')\le 0$.
Analogously, $(f''-P_Bf'',P_Bf'-P_Bf'')\le 0$.
Thus,
$$
\align
&(f'-f'',P_Bf'-P_Bf'') \\
&=||P_Bf'-P_Bf''||^2-(f'-P_Bf',P_Bf''-P_Bf')
-(f''-P_Bf'',P_Bf'-P_Bf'')\\
&\ge ||P_Bf'-P_Bf''||^2.
\endalign
$$
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 $XYZ$ and $YZT$ are not less than
$\frac\pi2$,
then $|YZ|\le|XT|$.
\medskip
Thus, $P_Bf$ depends continuously on $f$ (actually we proved that
$P_B\in Lip1$). Let's now show that it also depends continuously
on $B$.
\ul{Lemma 2.}
Let $B_H=\{x\in H:||x||\le1\}$ be the unit ball in $H$. Let $\d>0$, $f\in H$.
If two standard sets $B$ and $B'$ satisfy $P_{B'}f\in B+\d B_H$
and $P_Bf\in B'+\d B_H$ (i.e. $P_{B'}f$ is not more than $\d$
distant from $B$ and vice versa), then
$$
||P_Bf-P_{B'}f||\le\sqrt{2||f||\d}
$$
\ul{Proof}
Let $a=||f-P_Bf||$, $b=||f-P_{B'}f||$. Suppose, for definiteness,
that $a\ge b$. 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 $P_Bf$.
Let $H_+$ be the half-space containing $f$ and $H_-$ be the half-space,
containing $B$. As $a\ge b$, we have $P_{B'}f\in H_+$. Let
$u= ||P_Bf-P_{B'}f||$, $v$ 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 $\d$, and as $b\le a$,
we get $u\le\sqrt{2a\d}$. Since $0\in B$, we have
$a=||f-P_Bf||\le||f-0||=||f||$ and the conclusion of Lemma 2 follows
immediately.
\ul{The Bang functional $\Phi_B$.}
Let $B$ be a standard set in $H$. Consider the functional $\Phi_B:H\to
\Bbb R$ defined by
$$
\Phi_B(f)=||f||^2-||f-P_Bf||^2=2(f,P_Bf)-||P_Bf||^2\qquad(f\in H).
$$
The functional $\Phi_B$ has the following remarkable
\ul{Property:}
$$
\Phi(f)-\Phi(f')\le2(P_Bf,f-f')-||P_Bf-P_Bf'||^2.
$$
\ul{Proof:}
$$
\align
\Phi(f)-\Phi(f')&=2(f,P_Bf)-2(f',P_Bf')-||P_Bf||^2+||P_Bf'||^2\\
&=2(P_Bf,f-f')+2(f',P_Bf-P_Bf')-(P_Bf+P_Bf',P_Bf-P_Bf')\\
&=2(P_Bf,f-f')+(2f'-P_Bf-P_Bf',P_Bf-P_Bf')\\
&=2(P_Bf,f-f')+2(f'-P_Bf',P_Bf-P_Bf')-||P_Bf-P_Bf'||^2.
\endalign
$$
It remains only to note that the second term is nonpositive.
\ul{Observation:}
Let $\{\psi_j\}$ 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 $\e=\{\e_j\}$ where $\e_j=1$ or $-1$ the series
$\sum_j\e_ja_j\psi_j$ converges in $H$ to some element $f_\e$.
Consider the cortege $\bar\e$ for which $\Phi_B(f_\e)$ attains its maximal
value (as $\Phi_B$ is continuous and as convergence of all the series
$\sum_j\e_ja_j\psi_j$ implies their uniform convergence in $H$ with
respect to the choice of signs $\e_j$, the cortege $\bar\e$ exists
and can be constructed by diagonal process, say).
Let $f=f_{\bar\e}$ be the corresponding element of $H$. Fix some $j$
consider the series in which $\bar\e_j$ is replaced by $-\bar\e_j$. Let
$f'_j=f-2\bar\e_ja_j\psi_j$ be the sum of this series.
Then
$$
|(P_Bf,\psi_j)|\ge\frac{||P_Bf-P_Bf_j'||^2}{4a_j}
$$
\ul{Proof:}
We have
$
0\le\Phi_B(f)-\Phi_B(f'_j)\le 2(P_Bf,f-f'_j)-||P_Bf-P_Bf'_j||^2
$
which results in
$
2(P_Bf, f-f'_j)\ge ||P_Bf-P_Bf'_j||^2
$
and, after recalling that $f-f'_j=2\bar\e_ja_j\psi_j$, in
$$
\bar\e_j(P_Bf,\psi_j)\ge\frac{||P_Bf-P_Bf'_j||^2}{4a_j}.
$$
It remains to note that $|\bar\e_j|=1$.
\medskip
Let now $X=L^\infty(T)$, $H=L^2(T)$ where $T$
is some measure space with
probability measure $\mu$. Let $\psi_j$ be
a system of vectors in $H$ satisfying
$||\sum_jc_j\psi_j||\le \sqrt{\sum_jc_j^2}$ for every cortege of
coefficients $c_j\in\Bbb R$ and obvious necessary condition
$\int_T|\psi_j|d\mu\ge\b>0$. At last, let $a_j>0$ satisfy $\sum_ja_j^2=1$.
Then for every cortege $\e$ we have the corresponding series convergent
in $L^2(T)$ to the sum $f_\e$ of norm $||f_\e||_H\le 1$.
Let $s>0$, $B=sB_{L^\infty(T)}$.
The projection $P_Bf$ is easy to compute for every $f\in H$. Namely,
$$
(P_Bf)(t)=\left\{
\aligned
-s, &\text{ if } f(t)\le-s;\\
f(t), &\text{ if } -s\le f(t)\le s;\\
s, &\text{ if } f(t)\ge s
\endaligned
\right.
\qquad\quad (t\in T).
$$
Thus, (for the same $f$ and $f'$ as above), $||P_Bf-P_Bf'||^2\ge
\int_E|f(t)-f'(t)|^2d\mu(t)=4a_j^2\int_E\psi_j^2d\mu$, where
$E=\{t\in T: |f(t)|\le s, |f'(t)|\le s\}$.
The measure of the complement $T\setminus E$ of the set $E$ can
be estimated from above by $\frac{2}{s^2}$ (just from Tschebyshev
inequality). On the other hand, if $\mu(T\setminus E)\le\frac {\b^2}4$,
then
$$
\int_{T\setminus E}|\psi_j|d\mu\le
\mu(T\setminus E)^{\frac12}(\int_T\psi_j^2d\mu)^{\frac12}\le\frac\b2
$$
and thereby
$$
\int_E\psi_j^2d\mu\ge\left(\int_E|\psi_j|d\mu\right)^2\ge\frac{\b^2}4
$$
Thus, if we put $s=\frac{4}\b$, say, we will have
$$
|(P_Bf,\psi_j)|\ge\frac{\b^2}4a_j
$$
for every $j$. Then the function $F=\frac 4{\b^2}P_Bf$ has $L^\infty$-norm
$||F||_{L^\infty(T)}\le\frac {16}{\b^3}$ and its Fourier coefficients
with respect to $\psi_j$ are greater than $a_j$ in absolute value.
So, $L^\infty(T)$ is large in $L^2(T)$.
\ul{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 $L^\infty$) 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 $\{\psi_j\}$ is finite, all we need is to take the
new function $F'\in C(T)$ 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 $\dots$ 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 $\{\psi_j\}$
(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 $B=sB_{L^\infty(T)}$ with
$s=\frac {4}\b$ as before, the sequence of numbers $a_j>0$ satisfying
$\sum_ja_j^2=1$ and the corresponding family $\{f_\e=\sum_j\e_ja_j\psi_j:
\e_j=1\text{ or }-1\}$. Note that the family $\{f_\e\}$ is compact and
thereby the set $\{P_Bf_\e\}$ is compact as well (both in topology of
$L^2(T)$, of course). As $sB_{C(T)}$ is dense in $B=sB_{L^\infty(T)}$
(again, in $L^2$-sense),
we can find for every $\d>0$ a finite set of elements
$g_1,\dots,g_{N(\d)}\in sB_{C(T)}$
such that for every cortege $\e$ the element $P_Bf_\e$ is not more
than $\d$ distant from one of $g_m$. Let $B(\d)$ be the convex hull
of the origin and the points $g_m$. Then $B(\d)$ is a standard set
for every $\d>0$ and $B(\d)\subset sB_{C(T)}\subset B$
(the main advantage of $B(\d)$ compared to $sB_{C(T)}$ is that
$B(\d)$ is closed in topology of $L^2(T)$). Besides, every element
$P_Bf_\e$ is not more than $\d$ distant from $B(\d)$.
Fix some $\e$ and consider again the pair of functions $f=f_\e$
and $f'_j=f-2\e_ja_j\psi_j$.
We have
$$
||P_{B(\d)}f-P_{B(\d)}f'_j||\ge
||P_{B}f-P_{B}f'_j||-||P_{B(\d)}f-P_{B}f||-||P_{B(\d)}f'_j-P_{B}f'_j||.
$$
But, as we saw above,
the first term on the right is at least $\b a_j$ while lemma 2 together
with the estimates $||f||,||f'_j||\le 1$ implies that both other tems
do not exceed $\sqrt{2\d}$. Thus, if $\d\le\frac{\b^2a_j^2}{32}$,
we still have
$$
||P_{B(\d)}f-P_{B(\d)}f'_j||\ge\frac\b2a_j.
$$
Let us now consider some sequence $\d_k\to0$ and define the functional
$\Phi:H\to \Bbb R$ to maximize by
$$
\Phi(f)=\sum_{k=1}^\infty\frac1{k(k+1)}\Phi_{B(\d_k)}(f).
$$
Again, let $\bar\e$ be the cortege for which $\Phi$ attains its
maximal value. For the functions $f=f_{\bar\e}$ and $f'_j=f-2\bar\e_j
a_j\psi_j$ we get
$$
0\le\Phi(f)-\Phi(f'_j)\le\sum_{k=1}^\infty\frac1{k(k+1)}[2(P_{B(\d_k)}f,
f-f'_j)-||P_{B(\d)}f-P_{B(\d)}f'_j||^2]
$$
Thus, for the function $F=\sum_k\frac1{k(k+1)}P_{B(\d)}f$ (which clearly
is in $C(T)$) we have
$$
\bar\e_j(F,\psi_j)\ge\frac1{4a_j}\sum_{k=1}^\infty\frac1{k(k+1)}
||P_{B(\d)}f-P_{B(\d)}f'_j||^2.
$$
But
$$
\sum_{k=1}^\infty\frac1{k(k+1)}
||P_{B(\d)}f-P_{B(\d)}f'_j||^2\ge\sum_{\tsize{k:\d_k\le\frac{\b^2a_j^2}{32}}}
\frac1{k(k+1)}\frac{\b^2}4a_j^2=
\frac{\b^2a_j^2}{4k(a_j)},
$$
where for every $a>0$ the index $k(a)$ is defined as the first index
$k$ for which $ \d_k\le\frac{\b^2a_j^2}{32}$.
This results in
$$
|(F,\psi_j)|\ge\frac{\b^2}{16}\frac{a_j}{k(a_j)}
$$
for every $j$.
As the sequence $\d_k$ can tend to $0$ as fast as we desire, $k(a)$ is
just an arbitrary decreasing finction which has values $1,2,3,\dots$
and tends to infinity as $a\to0$. Therefore the sequence
$\{\frac{a_j}{k(a_j)}\}$ is nothing but an arbitrary sequence $b_j$
of positive numbers satisfying
$\sum_jb_j^2<1$. Thus, $C(T)$ is large and we have done.
\ul{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 $L^\infty$-norms of partial sums) is large under the same assumptions
about $\psi_j$ 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:
\ul{Proposition:}
Suppose that a sequence of numbers $a_j\in(0,\frac12)$, say,
satisfies the condition
$$
\sum_ja_j^2|\log a_j|^\gamma<\infty\text{\qquad with some }\g>2.
$$
Then there is a function $F\in U(\T)$ for which $|(F,\psi_j)|\ge a_j$
for every $j$.
\ul{Proof:}
Let us do the same as for the case $X=C(T)$. The main difference
is that now we can approximate an element $P_Bf_\e\in B=sB_{L^\infty(\T)}$
by an element $g\in U(\T)$ 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
\ul{Kislyakov correction theorem:}
If $||f||_{L^\infty(\T)}\le1$, then for every $\d\in(0,\frac12)$, say,
there exists $g\in U(\T)$ such that $||f-g||_{L^2(\T)}\le\d$ while
$||g||_{U(\T)}\le C\log\frac1\d$
($C>0$ is some absolute constant).
\medskip
We can now repeat the construction of standard sets $B(\d)$ using
the Kislyakov theorem instead of trivial approximation of
bounded functions by continuous ones.
For every $\d>0$ we see that $B(\d)\subset B+\d B_{L^2(\T)}$ and
the $U(\T)$-norms of all elements in $B(\d)$ do not exceed $Cs\log\frac s\d$.
It allows to repeat all the estimates except that for the norm of
the element $F$ constructed.
Now, to get the series $\sum_k\frac1{k(k+1)}P_{B(\d)}f$ convergent
in $U(\T)$ we have to demand $\sum_k\frac1{k(k+1)}\log\frac1{\d_k}<\infty$.
We will just put $\d_k=\exp\{-k^{\frac2\g}\}$ 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 $k(a)\approx |\log a|^{\frac\g2}$
for small $a$ and we conclude that
for every sequence of numbers $a_j\in(0,\frac12)$ satisfying $\sum_ja_j^2=1$
there is a function $F\in U(\T)$ such that for every $j$
$$
|(F,\psi_j)|\ge\frac{a_j}{|\log a_j|^{\frac\g2}},
$$
which is equivalent to the Proposition to prove.
%\endcomment
\centerline{\bf THE BANG SOLUTION OF THE COEFFICIENT PROBLEM -- 4}
\centerline{\bf How Bang solved the plank problem.}
Let $B\subset\R^n$ 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 $a_j\ed\frac{h_j}{2}$ and consider the
vectors $\psi_j$ of length $a_j$ orthogonal to the boundary hyperplanes of
the corresponding strips $S_j$.
Bang's solution consists of two independent steps:
\ul{Lemma 1:}
{\it If $\psi$ is a vector of length $a<\frac H2$, then the intersection
$(B-\psi)\cap (B+\psi)$ contains a homotetic image of $B$ with coefficient
$\frac{H-2a}{H}$ }
\ul{Corollary:}
{\it If $\sum_jh_j0$ and therefore is not empty.
Every point of this intersection can be taken as $x_0$.
\ul{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 $\sum_ja_j$ in the plank problem and $\sum_ja_j^2$ in the
coefficient problem comes from: in the plank problem we deal with
completely arbitrary
vectors $\psi_j$ 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.
\ul{Proof of lemma 2:}
Let $p_k$ be any point of the ``middle'' hyperplane of the strip $S_k$.
Then
$$
S_k=\{x\in\R^n: |(x-p_k,\psi_k)|0$. Despite the paper has
got quite famous during that year and despite all my own efforts, this has not
happened yet. Moreover, even a simpler question about ``largeness'' of
$X=H^\infty(\Bbb D)$ in $H=H^2(\Bbb D)$ remains unanswered.
Still I've got several interesting remarks from various people. Two most
interesting of them were the following.
First, Keith Ball pointed out to me that the ``coefficient problem'' can be
solved completely if one is interested in majorizing a $\ell^1$ sequence
$\{a_j\}$ instead of $\ell^2$ that. Namely, in 1991 he proved the following
\ul{Theorem.}
{\it Let $X$ be any Banach
space, $\psi_j\in X^*$ be any family of bounded linear
functionals on $X$ such that $\|\psi_j\|\ci{X^*}=1$ for every $j$. Then for
any sequence
$\{a_j\}$ of positive numbers satisfying $\sum_j a_j<1$ there exists $f\in X$
such that $\|f\|\ci X\le 1$ and $|(f,\psi_j)|\ge a_j$ for every j.}
\medskip
\noindent The proof of the theorem can be found in [3].
The second remark is due to
Francoise Lust-Piquard. She noticed that the Bang approach works also in a
``non-commutative'' setting. Among other results her forthcoming paper [4]
contains the following
\ul{Proposition.}
{\it Let $\{a_{ij}\}_{i,j=1}^\infty$ be a matrix with positive entries
such that
$$
\sum_{j=1}^\infty a_{ij}^2\le 1\text{ for every }i\quad\text{and}\quad
\sum_{i=1}^\infty a_{ij}^2\le 1\text{ for every }j.
$$
Then there exists a bounded operator $A:\ell^2\to\ell^2$ of norm $\|A\|\le
10$, say, such that its matrix satisfies $|A_{ij}|\ge a_{ij}$ for every
$i,j$.}
\medskip
The above condition is obviously necessary for existence of such an operator
of norm~1.
\bigskip
\leftline{\bf Literature cited}
\medskip
[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
Sér. 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.
\bye
\bye
**