7

2

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.

flag
1 
 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 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. – ET Oct 15 at 13:43
The post had a misprint. Corrected now. – ET Oct 15 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 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 at 21:16
show 3 more comments

1 Answer

3

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 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. :)

link|flag
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 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 at 3:55
You say this is a classical approach. Can you refer me to Bernstein's and Remez's relevant papers? – ET Oct 23 at 7:53
No. This is my weak point: I remember the proofs well, I remember names with some degree of accuracy, I do not remember references at all. So, I'll have to do exactly the same google or mathscinet search to answer this question as you can do yourself spending the same amount of time :(. – fedja Oct 23 at 12:49
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. – ET Oct 23 at 21:14
show 7 more comments

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.