Sup norm of iterated convolution on $\mathbb{R}$ Let $f\in L^\infty(\mathbb{R})$ be a bounded function satisfying $0\leq f\leq 1$ and
$\int f = 1$.  I am curious whether the bound
$$
\| f^{\ast k} \|_{L^\infty} \leq C k^{-1/2}
$$
holds for the $k$-fold iterated convolution $f^{\ast k}$.  This seems like a continuous version of the Littlewood-Offord problem, but I don't know how to make the connection precise.
 A: The following proves the claim for a constant $C = C(f)$ depending on the function $f$
under consideration.
I do not know whether the claim holds with a constant that is independent of $f$.
Note that it is enough to prove the claim for large $k$.
Let us write $\phi(\xi) = \int_{\mathbb{R}} e^{i x \xi} f(x) \, d x$
for the characteristic function of the probability density function $f$,
noting that $\phi(\xi) = \widehat{f}(- \xi / (2 \pi))$, with my preferred
normalization of the Fourier transform
$\mathcal{F} f (\xi) = \widehat{f}(\xi) = \int_{\mathbb{R}} e^{-2 \pi i x \xi} f(x) \, d x$.
Then, using Fourier inversion $f^{\ast k} = \mathcal{F}^{-1} \widehat{f^{\ast k}}$,
the fact that $\| \mathcal{F}^{-1} g \|_{L^\infty} \leq \| g \|_{L^1}$,
the convolution theorem, and Plancherel's theorem, we see
$$
  \| f^{\ast k} \|_{L^\infty}
  \leq \| \widehat{f^{\ast k}} \|_{L^1}
  = \| (\widehat{f})^k \|_{L^1}
  .
$$
Thus, it is enough to prove that $\| (\widehat{f})^k \|_{L^1} \lesssim k^{-1/2}$, as $k \to \infty$.
To see this, we use the final corollary from the paper
Lower and upper bounds for characteristic functions,
which states the following:

Define the concentration function of $f$ as
$$
    Q_f (L)
    := \sup_{a \in \mathbb{R}}
         \int_{a}^{a + L}
           f(x)
         \, d x
    .
  $$
Then, for any $L > 0$, the following inequalities hold:
$$
    |\phi(t)| \leq 1 - \frac{[Q_f (L)]^3}{3 \pi^2} t^2
    \quad \text{for } |t| \leq \pi / L
  $$
and
$$
    |\phi(t)| \leq 1 - \frac{[Q_f (L)]^3}{3 L^2}
    \quad \text{for } |t| > \pi / L
    .
  $$

In the following, I choose $L = 1$---one can probably optimize the result by using a different choice.
By applying the above result and recalling that $\phi(\xi) = \widehat{f}(- \xi / (2 \pi))$,
we see then that there exists a constant $c = c(f) > 0$ (in fact, $c = [Q_f(1)]^3 / 3$ will do)
satisfying
\begin{equation}
  |\widehat{f}(\xi)| \leq 1 - c \xi^2
  \quad \text{for } |\xi| \leq \frac{1}{2}
  \qquad \text{and} \qquad
  |\widehat{f}(\xi)| \leq 1 - c
  \quad \text{for } |\xi| > \frac{1}{2}
  .
  \tag{$\ast$}
\end{equation}
Note that $c \leq 1 / 3 < 1$.
Therefore, we see on the one hand that
\begin{align*}
  \int_{-1/2}^{1/2}
    \bigl|[\widehat{f}(\xi)]^k\bigr|
  \, d \xi
  & \leq \int_{-1/2}^{1/2}
           (1 - [\sqrt{c} \xi]^2)^k
         \, d \xi \\
  & = c^{-1/2}
      \int_{-\sqrt{c}/2}^{\sqrt{c}/2}
        (1 - \eta^2)^k
      \, d \eta \\
  & \leq 2 c^{-1/2}
        \int_{0}^{1}
          (1 - \eta^2)^k
        \, d \eta \\
  & = c^{-1/2}
      \int_0^1
        (1 - x)^k
        x^{-1/2}
      \, d x
    = c^{-1/2} \, B(k+1, \tfrac{1}{2})
\end{align*}
with the usual Beta function $B(a,b) = \int_0^1 t^{a-1} (1-t)^{b-1} \, d t$.
Now, Wikipedia tells me that
there exists a constant $c' > 0$ satisfying (at least for large $k$) that
$B(k+1, \tfrac{1}{2}) \leq c' \cdot (k+1)^{-1/2} \leq c' \cdot k^{-1/2}$.
Thus, the "low frequency part" is under control.
For the "large frequency part", we note that $f \in L^1 \cap L^\infty \subset L^2$
and thus also $\widehat{f} \in L^2$ and therefore $[\widehat{f}]^2 \in L^1$
(in fact, $\| \widehat{f} \|_{L^2} = \| f \|_{L^2} \leq 1$).
Furthermore, note that there exists a suitable constant $c'' > 0$ satisfying
for $k$ large enough that $|\widehat{f}(\xi)|^{k-2} \leq (1 - c)^{k-2} \leq c'' k^{-1/2}$
for $|\xi| > \frac{1}{2}$; here we again used the estimate $(\ast)$ for $\widehat{f}$.
Overall, we thus see
$$
  \int_{|\xi| > \frac{1}{2}}
    |\widehat{f}(\xi)|^k
  \, d \xi
  \leq c''
       \cdot k^{-1/2}
       \cdot \int_{|\xi| > \frac{1}{2}}
               |\widehat{f}(\xi)|^2
             \, d \xi
  \leq c''' \cdot k^{-1/2}
  .
$$
Combining the two parts, and recalling the estimate from the beginning of the proof, we are done.
Remark: The main estimate on which everything relies (apart from elementary Fourier analysis)
is the estimate $|\phi(t)| \leq 1 - c t^2$ for the characteristic function of a bounded probability
density.
I didn't know of this estimate before and think that it is quite interesting.
