# A question in Fourier analysis

I recently came across this problem:

Let $T=\mathbb R /2 \pi \mathbb Z$ the circle, with its proabability Haar measure $\mu$. Any integrable function $f : T \rightarrow \mathbb C$ has Fourier coefficients $c_n(f)= \int_T f(t) e^{-int} \frac{dt}{2\pi}$, and we define $$||f||_{A(T)} = \sum_{n \in \mathbb Z} |c_n(f)| \leq + \infty$$ its algebra norm. (This definition is standard)

Now let $E$ be a measurable subset of $T$. I want to compute, or estimate, $$\alpha(E) = \inf \{ ||f||_{A(T)}\ | \ f : T \rightarrow \mathbb R, \ f_{|E} \leq 0, \int_T f = 1 \}$$

An easy lower bound for $\alpha(E)$ is $\alpha(E) \geq \frac{1}{1-\mu(E)}$, but Id like to know as many thing as possible about $\alpha(E)$. For example, here is a basic question:

When $E$ is an interval such that $\mu(E) \rightarrow 1$, is that lower bound sharp?

Also, has this problem, or something related, been studied ? Does $\alpha(E)$ have a name ?

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Out of curiosity: is it important for your purposes that $f\leq 0$ on $E$, rather than $f=0$ on $E$? –  Yemon Choi Jun 22 '13 at 3:44
@Yemon: Yes, kind of. I mean, I am mostly interested in upper bound on $\alpha(E)$, so any such bound for the variant $\alpha'(E)$ you suggest, where the inf is taken over function which are zero outside $E$, obviously give a bound that interests me for $\alpha(E)$. But in my problem, I have no need to impose the condition $f_{|E}=0$, so if I can get better bound without that, the better. –  Joël Jun 24 '13 at 14:34
Yemom and Noam, thanks for both your great answers. I accept Noam's as it gives a better upper bound, discusses also lower bounds, and gives plenty of ideas but already Yemon's one was very helpful. –  Joël Jun 24 '13 at 14:38
By the way, from the fact that neither Yemon nor Noam nor anyone else here mentioned anything related to this, it seems more likely than before than this notion of $\alpha(E)$ (or its variant $\alpha'(E)$ defined with the condition $f_{|E}=0$) has no official name and was not extensively studied. Or am I wrong? –  Joël Jun 24 '13 at 14:44
Joel: I have not seen either $\alpha$ or $\alpha'$ before, but somehow $\alpha'$ "looks closer to familiar settings" than $\alpha$ - something to do with finding elements of an ideal that satisfy a linear constraint. If I see anything I will leave a comment or an update –  Yemon Choi Jun 25 '13 at 5:43

Yemon Choi (henceforth "YC") answered the question to within a constant factor with a function $f_\delta$ showing that $\alpha(E_\delta) \leq 2/\delta$ where $E_\delta$ is an interval of measure $1-\delta$ (YC's notation). That's optimal asymptotically $-$ and even exactly if $2/\delta \in {\bf Z}$ $-$ for functions $f$ with nonnegative Fourier coefficients $c_n(\phantom.f)$. But without this additional hypothesis on the $c_n(\phantom.f)$ the constant can be improved somewhat: we give another piecewise linear function (still vanishing on $E_\delta$) that reduces the constant from $2$ to below $1.9$, and that's surely not optimal either. But it certainly can't go all the way down to Joël's lower bound of $1/\delta$: the constant must be at least $3/2$, and I think it must be at least $\pi/2$ if $f=0$ on $E_\delta$ as YC suggested.

If $\| \phantom. f \|_{A(T)} \lt \infty$ then $\sum_n c_n(\phantom.f) e^{int}$ converges absolutely to a continuous function with Fourier coefficients $c_n$, so we may assume $f$ is continuous. If moreover each $c_n(\phantom.f) \geq 0$ then $\| \phantom. f \|_{A(T)} = \sum_{n\in{\bf Z}} c_n = f(0)$. Now suppose $f(t) \leq 0$ for $t \in E_\delta$, and let $N$ be the integer $\lfloor 2/\delta \rfloor$. We have $f(m/N) \leq 0$ for all integers $m \not\equiv 0 \bmod N$. Therefore $$f(0) \geq \sum_{m \phantom. \bmod \phantom. N} f(m/N) = N \sum_{N\phantom.|\phantom.n} c_n(\phantom.f) \geq N c_0(\phantom.f) = N$$ because $c_0 = \int_T f \phantom. d\mu = 1$; this proves our claim.

To give an example with smaller $\| \phantom. f \|_{A(T)}$, consider the following generalization: YC's function $f_\delta$ is the self-convolution of uniform meausre on an interval of length $\delta/2$; for $0 \lt b \lt 1$ define $f_{\delta,b}$ to be the convolution of uniform measures on intervals of length $b\delta$ and $(1-b)\delta$. Then $$c_n(\phantom.f_{\delta,b}) = \frac{\sin(\pi b \delta n)}{\pi b \delta n} \frac{\sin(\pi (1-b) \delta n)}{\pi (1-b) \delta n}.$$ Hence $\delta \sum_n \left| c_n(\phantom.f_{\delta,b}) \right|$ is a Riemann sum for $$\int_{-\infty}^\infty \left| \frac{\sin(\pi b s)}{\pi b s} \frac{\sin(\pi (1-b) s)}{\pi (1-b) s} \right| \phantom. ds.$$ Numerical calculation shows this integral decreasing from $2$ at $b=1/2$ (YC's function) down to about 1.9 at $b \sim 0.38$, then up slightly and back down to just below $1.9$ at $b \sim 0.29$ before increasing and crossing $2$ at $b\sim 0.14$ (it must blow up at $b \rightarrow 0$). We do a bit better by averaging $f_{\delta,0.29}$ with $f_{\delta,0.38}$, getting just below $1.875 = \frac{15}{8}$.

This kind of analysis suggests that the asymptotic value of $\delta \cdot \alpha(E_\delta)$ should be the infimum, call it $\alpha_0$, of $\int_{-\infty}^\infty \bigl|\phantom.\hat f(s)\bigr| ds$ over functions $f: {\bf R} \rightarrow {\bf R}$ such that both $f$ and its Fourier transform $$\hat f(s) = \int_{-\infty}^\infty e^{-2\pi i s t} f(t) dt$$ are absolutely integrable and satisfy $\hat f(0) = 1$ and $f(t) \leq 0$ for all $t$ with $|t| \geq 1/2$. I don't quite have a proof of this, but all examples and bounds so far translate immediately to $\alpha_0$. For example, if we also assume that $\hat f(s) \geq 0$ for all $s$ then $f(0) \geq 2$ follows from Poisson summation: $$f(0) \geq \sum_{m \in {\bf Z}} f(m/2) = 2 \sum_{s \in {\bf Z}} f(2n) \geq 2\hat f(0) = 2.$$ This is the 1-dimensional case of the inequality in Henry Cohn's thesis on upper bounds of sphere-packing densities.

The dictionary between bounds on $\delta \alpha(E_\delta)$ as $\delta \rightarrow 0$ and bounds on $\alpha_0$ also works for the following improved lower bound, which I'll give only in the $\alpha_0$ setting. Here Joël's "easy lower bound" is $\alpha_0 \geq 1$, proved by noting that $$1 = \int_{-\infty}^\infty f(t) dt \leq \int_{-1/2}^{1/2} f(t) dt$$ implies $f(t) \geq 1$ for some $t \in [-1/2, +1/2]$, and thus $\bigl\| \phantom. \hat f \bigr\|_1 \geq 1$. But clearly equality cannot hold here. To exploit this, we write

$$\int_{-1/2}^{1/2} f(t) dt = \langle \phantom. f, \chi \rangle = \langle \phantom. \hat f, \hat\chi \rangle \leq \bigl\| \phantom. \hat f \bigr\|_1 \bigl\| \hat\chi \bigr\|_\infty$$ where $\chi$ is the characteristic function of $[-1/2,1/2]$. Now $\bigl\| \hat\chi \bigr\|_\infty = 1$, but only because $\hat\chi(0) = 1$. So we can improve the estimate by writing $\langle \phantom. f, \chi \rangle \leq \langle \phantom. f, \psi \rangle$ for any function (or even any measure) $\psi$ such that $\hat\chi - \hat\psi$ is nonnegative and supported outside $(-1/2, 1/2)$; if $\bigl\| \hat\psi \bigr\|_\infty \lt 1$ then $1 \bigl/ \bigl\| \hat\psi \bigr\|_\infty \bigr.$ is an improved lower bound on $\alpha_0$. A choice that gives $\alpha_0 \geq 3/2$ is $$\hat\psi(s) = \hat\chi(s) - \frac13 \cos(\pi s) = \frac{\sin \pi s}{\pi s} - \frac13 \cos(\pi s) = \frac23 - \frac1{180} (\pi s)^4 + - \cdots,$$ so $\langle \phantom. f, \psi \rangle = \langle \phantom. f, \chi \rangle - \frac16\bigl( \phantom.f(1/2) + f(-1/2)\bigr) \geq \langle \phantom. f, \chi \rangle \geq 1$; then $\bigl\| \hat\psi \bigr\|_\infty \lt 2/3$, so $\alpha_0 \geq 3/2$ as claimed.

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Just to get the ball rolling: for any $\delta \in(0,1)$, the function $f_\delta(t)$ which is supported on the interval $[-\delta\pi,\delta\pi]$ and looks like $1- (\delta\pi)^{-1}|t|$ inside it, has norm $1$ in $A(T)$ -- one either sees this using positive-definiteness or a trick with convolving characteristic functions in $L^2(T)$, or by a direct calculation one can show the Fourier coefficients are all positive.

[The following calculation was originally out by a factor of $2$ -- thanks to Noam Elkies for pointing this out.]

Clearly $\int_T f_\delta = \delta/2$. Therefore, by considering $2\delta^{-1}f_\delta$, and $E_\delta= T \setminus [-\delta\pi,\delta\pi]$, $$\alpha( E_\delta ) \leq 2\delta^{-1} = \frac{2}{1- \mu(E_\delta)}$$ which is a factor of $2$ out from the lower bound you state in your specific question. (My initial suspicion is that your lower bound is not exactly sharp, but I am hesitant to claim right now that the inequality above is equality.)

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That works within a factor of 2: $\int_T f_\delta$ is actually $\delta/2$, because $\mu([-\delta\pi,\delta\pi]) = \delta$ and the average of $f_\delta$ over that interval is $1/2$. So we must use $2\delta^{-1} f_\delta$, and deduce the bound $\alpha(E_\delta) \leq 2\delta^{-1}$. –  Noam D. Elkies Jun 21 '13 at 22:16
Thanks, Noam. I've edited with the correction. –  Yemon Choi Jun 21 '13 at 22:28