An optimization problem: $\Phi(0)$, $\widehat{\Phi}(0)$, $\Phi$ a majorant (This is a problem that arose from my own answer to Mean value theorem for Dirichlet series - optimize? )
Let $\Phi:\mathbb{R}\to \mathbb{R}$ be a real-valued, symmetric, non-negative function such that $\Phi(t)\geq 1$ for $|t|\leq 1/2$, $\widehat{\Phi}$ has compact support, and $\widehat{\Phi}(t)$ is non-increasing for $t\geq 0$.
It follows immediately that $\Phi(0)\geq 1$ and $\widehat{\Phi}(0) = |\Phi|_1\geq 1$.
Questions:
(a) Can we choose $\Phi$ so that $\Phi(0)$ is equal to $1$, or arbitrarily close to $1$? How small can $\widehat{\Phi}(0)$ then be?
(b) Can we choose $\Phi$ so that $\widehat{\Phi}(0)$ is equal to $1$, or arbitrarily close to $1$? How small can $\Phi(0)$ then be?

Actually, the answer to (b) is most likely "no": $\Phi$ would have to be very close to the characteristic function $f=1_{[-1/2,1/2]}$ in $L^1$ norm, and, if $\Phi$ were bounded, it would follow that $\Phi$ has to be very  close to $f$ in $L^2$ norm - and so $\widehat{\Phi}$ would have to be very close to $\widehat{f}$ in $L^2$ norm; now, $\widehat{f}$ is $(\sin \pi t)/\pi t$, which is certainly not monotonic for $t\geq 0$.
So, let me ask a variant:
(b') How small can $\widehat{\Phi}(0)$ be? How small can $\Phi(0)$ then be?
 A: For starters we have $\,\widehat{\!\Phi\!}\,(0) \geq 2$.
Indeed by Poisson summation
$$
\sum_{m = -\infty}^\infty \Phi(m+1/2) =
\sum_{n = -\infty}^\infty (-1)^n \,\widehat{\!\Phi\!}\,(n).
$$
The LHS is at least $\Phi(-1/2) + \Phi(1/2) \geq 2$.
The RHS is at most $\,\widehat{\!\Phi\!}\,(0)$ because
$$
\sum_{n \neq 0} (-1)^n \,\widehat{\!\Phi\!}\,(n) =
-2\Bigl( 
 \bigl(\,\widehat{\!\Phi\!}\,(1) - \,\widehat{\!\Phi\!}\,(2)\bigr) 
   + \bigl(\,\widehat{\!\Phi\!}\,(3) - \,\widehat{\!\Phi\!}\,(4)\bigr) 
   + \cdots
\Bigr)
$$
and each term $\Phi(2k-1) - \Phi(2k)$ is nonnegative
because $\,\widehat{\!\Phi\!}\,(t)$ is non-increasing on $t \geq 0$.  Hence
$$
2 \leq \sum_{m = -\infty}^\infty \Phi(m+1/2)
  = \sum_{n = -\infty}^\infty (-1)^n \,\widehat{\!\Phi\!}\,(n)
  \leq \,\widehat{\!\Phi\!}\,(0),
$$
QED.
A: If $\Phi(0) = 1 + \epsilon$ then
$\,\widehat{\!\Phi\!}\,(t) \gg \epsilon^{-1/2}$,
even under the weaker assumption that
$\,\widehat{\!\Phi\!}\,(t) \geq 0$ for all $t$;
and this is best possible up to the constant factor.
Using Fourier inversion together with the inequality
$3 - 4 \cos \theta + \cos 2\theta \geq 0$ (all $\theta \in \bf R$)
we have $\Phi(2t) \geq 4 \Phi(t) - 3 \Phi(0)$ for all $t$.
Thus $\Phi(0) = 1 + \epsilon$
together with $\Phi(t) \geq 1$ for $|t| \leq 1/2$ implies
$\Phi(t) \geq 1 - 3\epsilon$ for $|t| \leq 1$, then
$\Phi(t) \geq 1 - 15\epsilon$ for $|t| \leq 2$, then
$\Phi(t) \geq 1 - 63\epsilon$ for $|t| \leq 4$, etc.
In particular if $4^{n+1} \epsilon \leq 1/2$ then
$\Phi(t) > 1/2$ for $|t| \leq 2^n$, so
$$
\,\widehat{\!\Phi\!}\,(0) = \int_{-\infty}^\infty \Phi(t) \, dt
\geq \int_{-2^n}^{2^n} \Phi(t) \, dt
\geq \int_{-2^n}^{2^n} \frac12 \, dt = 2^n.
$$
Using the largest such $n$ gives
$\,\widehat{\!\Phi\!}\,(t) \gg \epsilon^{-1/2}$ as claimed.
A Gaussian $\Phi(t) = (1+\epsilon) e^{-c \epsilon t^2}$
attains $\,\widehat{\!\Phi\!}\,(t) \sim C \epsilon^{-1/2}$
(with $\,\widehat{\!\Phi\!}\,(t)$ not of compact support but
still decreasing to zero on $0 \leq t \to \infty$).
