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\def \st #1 \\ #2 \endst{\medskip\leftline{\bf\underbar{#1.}}\nopagebreak
\smallskip\nopagebreak
{#2}\medskip}
\def\til#1{\centerline{\bf #1}}
\def\C{\Bbb C}
\def\T{\Bbb T}
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\redefine\ge{\geqslant}
\def\ed{\overset{\text{def}}\to=}
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\def\prf{\leftline{\bf Proof.}}
\def\pm{{\overset{{}_+}\to{{}_-}}}
\def\d{\delta}
\til{Complete version of the Turan lemma for trigonometric polynomials}
\til{on the unit circumference.}
\medskip
\til{\S 0. Introduction and preliminaries.}
Let $p(z)=\sum_{k=1}^nc_kz^{m_k}$ (~$c_k\in\C,\ \ m_1<\dotsy\}=\mu\{t\in[0,1]:\Phi(t)>y\}$ for every $y>0$)
it means that
$ \Phi(t)\ge\bigl(\frac{1-t}{14}\bigr)^{n-1}\Phi(0) $
and therefore for each set $E\subset\T$
$$
\multline
||p||_{L^0(E)}\ge\exp\biggl\{\frac1{\mu(E)}
\int_{1-\mu(E)}^1\log\,\Phi(t)\,dt\biggr\}
\\
\ge
\biggl(\frac{\mu(E)}{14e}\biggr)^{n-1}\Phi(0)=
\biggl(\frac{\mu(E)}{14e}\biggr)^{n-1}||p||_{L^\infty(T)}.
\endmultline
$$
Thus, for every $q\in[0,+\infty]$
$$
||p||_{L^q(\T)}\le\biggl\{\frac{14e}{\mu(E)}\biggr\}^{n-1}
||p||_{L^q(E)}.
$$
As the example of the polynomial $p(z)=(z-1)^{n-1}$ and a small arc
$E\subset\T$ centered at $1$ shows, this estimate cannot be essentially
improved for sets $E$ of small measure even if we place $L^0(\T)$-norm
to the left hand side and $L^\infty(E)$-norm to the right hand that.
But for sets $E$ close to the whole circumference $\T$ one could expect
some inequality of the kind
$
||p||_{L^q(\T)}\le B(\mu(E))^{n-1}
||p||_{L^q(E)}.
$
where $B(\mu)\to 1$ when $\mu\to 1-$, while the factor given by Turan lemma
or by its generalization remains very far from 1 even for the whole
circumference and this drawback cannot be removed by a minor modification
of any proof known so far even for arcs $E$.
The main aim of this paper is to confirm this common
belief and to show that
for every $0\le q\le 2$ and each set $E\subset\T$ of measure
$\mu(E)\ge\frac13$ the estimate
$$
||p||_{L^q(\T)}\le e^{A(n-1)\mu(\E)}
||p||_{L^q(E)}\tag *
$$
holds with some absolute constant $A>0$ not depending on $q$ (by $\E$
we denote the complement $\T\setminus E$ of the set $E$).
It is not very difficult (but rather boring, so we aren't going to perform
it here) to check that there is an absolute constant $a>0$ such that
for each $q\in[0,2]$, $n\in\N$
and for every arc $E\subset\T$ of measure $\mu(E)
\ge\frac13$ one can find a polynomial $p(z)$ in not more than
$n$ characters
for which
$$
||p||_{L^q(\T)}\ge e^{a(n-1)\mu(\E)}
||p||_{L^q(E)}.
$$
(the polynomial in the example can always be taken of the form
$p(z)=\bigl(\frac{z^m-1}{z-1}\bigr)^r$ with some $m,r$ satisfying
$mr\le n$). So, the estimate above is the best possible one up to
the numerical value of the absolute constant involved. If $q>2$,
one can merely use the fact that
$
||p||_{L^\infty(\T)}\le \sqrt n
||p||_{L^2(\T)}
$
to get the same estimate for "not extremaly large" sets $E$, namely for
sets $E$ of measure $\mu(E)\le 1-\frac{\log n}n$. If the measure of
$\E$ is essentially less that $\frac{\log n}n$, the inequality (*)
is no longer true for $q>2$ and we'll return to the question what one
should replace it with in the end of the paper.
\medskip
\til{\S 1. The proof of (*) for q=2.}
\st Lemma 1 \\
For every four real numbers $0\le a\le A$ and $0\le b\le B$ the inequality
$$
A+B-2\sqrt{ab}\ge\biggl(\frac1{a+A}+\frac1{b+B}\biggr)
\sqrt{(A^2-a^2)(B^2-b^2)}
$$
holds.
\endst
\prf
Substituting $A=x(1+\alpha),\ a=x(1-\alpha)\quad B=y(1+\beta),\ b=y(1-\beta)$
we get
$$
x(1+\alpha)+y(1+\beta)\ge 2\sqrt{xy(1-\alpha)(1-\beta)}
+2(x+y)\sqrt{\alpha\beta}
$$
to prove. But
$$
\align
x(1-\beta)+y(1-\alpha)&\ge 2\sqrt{xy(1-\alpha)(1-\beta)}, \\
(x+y)(\alpha+\beta)&\ge 2(x+y)\sqrt{\alpha\beta}
\endalign
$$
and it remains only to add these inequalities to get the statement desired.
\st Lemma 2 \\
Let $f,g:\T\to\C$. Suppose that $f$ is even and $g$ is odd, i.e.
$f(-z)=f(z)$, $g(-z)=-g(z)$ for every $z\in\T$. For every set $H
\subset\T$ put $F(H)\ed\int_H|f|^2d\mu$, $G(H)\ed\int_H|g|^2d\mu$.
For every set $E\subset\T$ put $E^+\ed E\cup(-E)$, $E^-\ed E\cap(-E)$.
Then for the function $p=f+g$ one has
$$
\int_E|p|^2d\mu\ge
\sqrt{
\frac{F(E^+)}{F(\T)}\frac{F(E^-)}{F(\T)}\frac{G(E^+)}{G(\T)}\frac{G(E^-)}{G(\T)}
}\int_\T|p|^2d\mu.
$$
\endst
\prf
Let $E^*$ denote the "antisymmetric" part of $E$, i.e. $E^*\ed
\{z\in E:-z\notin E\}$. As $|f|^2$ and $|g|^2$ are both even, we get
$F(E^\pm)=F(E)\pm F(E^*)$; $G(E^\pm)=G(E)\pm F(E^*)$ and therefore
$$
\multline
\sqrt{
\frac{F(E^+)}{F(\T)}\frac{F(E^-)}{F(\T)}\frac{G(E^+)}{G(\T)}\frac{G(E^-)}{G(\T)}
}
\\
=\frac1{F(\T)G(\T)}\sqrt{
(F(E)^2-F(E^*)^2)(G(E)^2-G(E^*)^2)
}\endmultline
$$
Beside it, we have
$$
\multline
\int_E|p|^2d\mu=F(E)+G(E)+2Re\,\int_Ef\bar g\,d\mu
\\
=F(E)+G(E)+2Re\,\int_{E^*}f\bar g\,d\mu\ge
F(E)+G(E)-2\sqrt{F(E^*)G(E^*)}
\endmultline
$$
for $f\bar g$ is an odd function and therefore its integral over the
symmetric part of $E$ vanishes.
At last, $\int_\T|p|^2d\mu=F(\T)+G(\T)$.
Thus, it is enough to show that
$$
\multline
F(E)+G(E)-2\sqrt{F(E^*)G(E^*)}
\\
\ge\biggl(
\frac1{F(\T)}+\frac1{G(\T)}
\biggr)
\sqrt{
(F(E)^2-F(E^*)^2)(G(E)^2-G(E^*)^2)
}.
\endmultline
$$
But $F(\T)\ge F(E^+)=F(E)+F(E^*)$ and $G(\T)\ge G(E^+)=G(E)+G(E^*)$.
Substituting these trivial estimates into denominators, we see that
lemma 2 is a direct consequence of lemma 1 with $A=F(E)$, $a=F(E^*)$,
$B=G(E)$ and $b=G(E^*)$.
Let's now prove the inequality
$$
\int_E|p|^2d\mu\ge e^{-An\mu(\E)}\int_\T|p|^2d\mu
$$
in induction in $n$ (as above we assume that $p$ is a trigonometric
polynomial in $n$ characters and $\mu(E)\ge\frac13$).
Note that the estimate (*) for $q=2$ is equivalent to
$$
\int_E|p|^2d\mu\ge \mu(E)e^{-2A(n-1)\mu(\E)}\int_\T|p|^2d\mu
$$
which is always weaker except the trivial case $n=1$.
\leftline{\bf Inductional base.}
If $n=1$ we have
$$
\int_E|p|^2d\mu= \mu(E)\int_\T|p|^2d\mu \ge
e^{-An\mu(\E)}\int_\T|p|^2d\mu
$$
for every set $E$ of measure $\mu(E)\ge\frac13$
merely because $t\ge e^{-A(1-t)}$ when $t\ge\frac13$ and $A\ge 3$, say.
If $\frac13\le\mu(E)\le\frac23$, the estimate for "small" sets cited
above implies
$$
\multline
\int_E|p|^2d\mu\ge \mu(E)
\biggl\{\frac{\mu(E)}{14e}\biggr\}^{2(n-1)}
\int_\T|p|^2d\mu
\\
\ge
\biggl\{\frac{\mu(E)}{14e}\biggr\}^{2n}
\int_\T|p|^2d\mu \ge
e^{-An\mu(\E)}\int_\T|p|^2d\mu
\endmultline
$$
for every $n\in\N$ because
$
\left\{\frac t{14e}\right\}^{2n}\ge e^{-An(1-t)}
\text{ for $\frac13\le t\le\frac23$ and $A\ge 30$, say.}
$
\leftline{\bf Inductional step.}
Let $p$ be a polynomial in $n>1$ characters and $E$ be a subset of
$\T$ of measure $\mu(E)\ge\frac23$.
Decompose $p$ into the sum of its even and odd parts $f(z)\ed
\frac12(p(z)+p(-z))$ and $g(z)\ed\frac12(g(z)-g(-z))$. Without
loss of generality we may suppose that both $f$ and $g$ don't vanish
identically (otherwise we could take $z^2$ as a new variable; it is
really possible because the "worst" set $E$ (i.e. the set for which
the integral $\int_E|p|^2d\mu$ is the smallest one) is a level set of
$|p|$ and thereby has all the symmetries $|p|$ possesses).
Note that $f$ and $g$ are trigonometric polynomials in $n_1$ and $n_2$
characters, say, and that $n_1+n_2=n$. In notations of lemma 2, we get
$$
\int_E|p|^2d\mu\ge
\sqrt{
\frac{F(E^+)}{F(\T)}\frac{F(E^-)}{F(\T)}\frac{G(E^+)}{G(\T)}\frac{G(E^-)}{G(\T)}
}\int_\T|p|^2d\mu.
$$
Note that $n_1,n_22$, $\mu(E)>1-\frac{\log n}n$
remained. In this case I cannot prove as sharp estimates as for the
case $0\le q\le 2$, but I can aswer completely the following particular
question which seems to be of its own interest. The question is how
to describe the sets of pairs $\Omega=\{(n,\mu)\}$ for which there exists
a constant $B>1$ such that for every pair $(n,\mu)$ in $\Omega$ and
for every polynomial $p$ in $n$ characters and set $E\subset\T$
the inequality $\mu(\E)\le\mu$ implies
$
||p||_{L^q(I)}\le B
||p||_{L^q(E)}
$
.
The answer is that for $2