# Norm concentration of trigonometric polynomials - Uncertainty principle

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.

• One can get these sorts of bounds by using reproducing formulae coming from Littlewood-Paley theory, e.g. expressing $P_N = P_N * K_N$ where $K_N$ is a trig polynomial from -2N to 2N (say) with the coefficients from -N to N equal to 1, and smoothly decaying to zero outside of this. See e.g. Section 5 of my notes at math.ucla.edu/~tao/254a.1.01w/notes1.dvi for some examples of this (I do it on the real line rather than on the circle, but the general idea is the same). – Terry Tao Oct 14 '12 at 17:32
• Thanks for the comment. I got the general idea, though I'm still not able to get this sharp lower bound, not even on the real line. – Itay Oct 15 '12 at 13:43
• The post had a misprint. Corrected now. – Itay Oct 15 '12 at 14:14
• Do you want "for all $c>0$ there is $\epsilon>0$" or "there exist $c>0$ and $\epsilon>0$"? Even the former should be true but the latter is easier (at least for an interval). – Noam D. Elkies Oct 15 '12 at 19:54
• Ah, the bound you want is slightly tricky since the analogous statement for the p-adics is false, and so general Fourier-analytic methods are not sufficient. But I believe one can obtain this result by a compactness and contradiction argument using the fact (special to R) that there are no non-trivial compactly supported functions whose Fourier transform is again compactly supported. A bit more specifically, suppose the claim is false, take a sequence of counterexamples, and use tools such as the Rellich compactness theorem to extract a counterexample to the previous claim. – Terry Tao Oct 15 '12 at 21:16

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 $|P|<e^{10C}$ 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.
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 $C/N$.
• This looks promising but I think you're answering the question for the sup norm, not the $L^2$ norm which might be smaller by a factor of order $N^{1/2}$. Can the analysis be adapted from $L^\infty$ to $L^2$? – Noam D. Elkies Oct 20 '12 at 21:36
• Come on! I showed that $P$ is comparable to its maximum $M$ on a set of measure $10C/N$. Now just think a tiny bit. The integral of $|f|^2$ over any set E of measure $C/N$ is at most $CM^2/N$. However, the integral over the remaining part of the set of measure $10C/N$ is also at least $cM^2/N$, so the outside of E carries a noticeable part of the norm, as you wanted. That is the beauty of working with measurable sets instead of intervals: you actually get estimates for the distribution function and can switch from $L^\infty$ bounds to bounds in other norms at no cost whatsoever. – fedja Oct 23 '12 at 3:55
• I will give the search another try, thanks anyway :). On a more concrete note, can you please explain where does the bound $\left|P\right|<e^{10C}$ in $D$ come from? I must be missing something. – Itay Oct 23 '12 at 21:14