Norm concentration of trigonometric polynomials - Uncertainty principle - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:40:59Z http://mathoverflow.net/feeds/question/109624 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109624/norm-concentration-of-trigonometric-polynomials-uncertainty-principle Norm concentration of trigonometric polynomials - Uncertainty principle ET 2012-10-14T16:18:52Z 2012-10-21T13:31:21Z <p>Hi all, I am interested in the following question (which is quite similar to one I posed a long while ago): Let $P_{N}(t)=\underset{k=-N}{\overset{N}{\sum}}c_{k}e^{ikt}$ be a unit norm trigonometric polynomial, we look at it as a function of $L^{2}\left(\mathbb{T}\right)$. I'd like to find a direct proof to the fact that there exists $\varepsilon>0$ such that for every $N$ and every such polynomial $P_{N}$ we have $\underset{E}{\int}|{P_{N}( t)}|^{2}dt\leq1-\varepsilon$ whenever $E\subset\mathbb{T}$ of measure $|E|=\frac{c}{N}$, and $c>0$ is some absolute constant. I would be happy with a proof only in the case $E$ is an interval, if it is any different than the general case. To rephrase the statement; one cannot concentrate the norm of a trigonometric polynomial of degree $N$ on an interval (or any measurable set) of length (measure) of the order of magnitude $\frac{1}{N}$. Let me comment that there is a result by Nazarov which implies this but it is way too general for my purposes.</p> http://mathoverflow.net/questions/109624/norm-concentration-of-trigonometric-polynomials-uncertainty-principle/110067#110067 Answer by fedja for Norm concentration of trigonometric polynomials - Uncertainty principle fedja 2012-10-19T04:34:37Z 2012-10-19T04:34:37Z <p>Let's talk about algebraic polynomials (just multiply by $z^N$). Let $P$ be a polynomial of degree $N$. Let $\max|P|=1$ and assume that this maximum is attained at some point $p$. Take the disk $D$ of radius $10C/N$ centered at $p$. Note that $z^{-N}P(z)$ satisfies the maximum principle in the complement of the unit disk and $P(z)$ satisfies the maximum principle in the disk itself, so <code>$|P|&lt;e^{10C}$</code> in $D$. Rescale $D$ to the unit disk $\mathbb D$ and divide by $e^C$. You'll get a bounded by $1$ function $F$ that is $e^{-10C}$ at the center. </p> <p>Now it is the usual story about subharmonicity of the logarithm of an analytic function. Let $H$ be any closed set on the circular arc passing through the origin on which $F$ is very small. Suppose that the length of $H$ is $1/2$ or more. Consider the function $$U(z)=\int_H\log \frac{|z-w|}{|1-\bar wz|}d\ell(w)$$ It is easy to see that $U\ge -A$ in $\mathbb D$, harmonic in $\mathbb D\setminus H$, $U=0$ on the unit circumference, and $U(0)\le -a$ for some absolute $a,A>0$. If $|F|\le e^{-10CA/a}$ everywhere on $H$, then $\log|F|\le (10C/a)U$ on the boundary of $\mathbb D\setminus H$ and, thereby, $\log|F(0)|\le -10C$, which is not the case. Thus, the minimum of $|F|$ over every set $H$ of length $1/2$ or more is bounded from below. Coming back to the original problem, we see that $|P|$ is bounded from below by some constant depending on $C$ on at least half of the arc of length $10C/N$, so we have plenty of noticeable values outside any set of measure $C/N$.</p> <p>This trick is pretty old and goes back to Bernstein. There is also another approach due to Remez (moving zeroes and looking at the level sets). You are 100% right when saying that you do not need the Turan type bounds and the related fancy techniques for this problem. :) </p>