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 ?
 A: 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.
A: 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.)
