Mean value property with fixed radius

Note that

$$ \frac{1}{2}\int_{x-1}^{x+1}e^{as}ds = \frac{\sinh(a)}{a}e^{ax}. $$

Let $a+ib \in \mathbb{C}$ be a non-zero root of $\sinh(z)=z$. Then also $a-ib$ is a root and because the MVP is linear the function $f(x) = e^{ax}\cos(bx)$ has the mean value property.

Remains to prove that $\sinh(z)-z$ has non-zero roots. I present two proofs below. The second one (that I gave first) is a bit clunky but has the advantage that it gives some information about the distribution of the roots.

Proof 1: Let $f(z) = \sinh(z)-z$. Then $f$ has an essential singularity at $\infty$. By Picard's great theorem the function $f$ attains all values infinitely often with at most one exception. In particular at least one of $0$ and $2\pi i$ must be attained infinitely often. However, $f(z + 2\pi i) = f(z) - 2\pi i$ and so both cases imply that $f$ has infinitely many roots.

Proof 2: I'll show that $|\sinh(z)|=|z|$ holds along some curve in $\mathbb{C}$ extending to $\infty$ and that $\sinh(z)/z$ must wind around the unit circle infinitely often along this curve.

If $z=x+iy$ then $2|\sinh(z)|^2 = \cosh(2x)-\cos(2y)$ and so $|\sinh(z)| = |z|$ exactly if $\cosh(2x)-2x^2=\cos(2y)+2y^2$. To show that for each $x$ there is a $y$ that satisfies this equation define the following functions:

$$ f(x) = \cosh(2x)-2x^2, \ g(x) = \cos(2x)+2x^2. $$

On $\mathbb{R}_{>0}$ both functions are strictly increasing and $f(x) > g(x)$. (The latter inequality can be checked from their power series.) In particular, for each $x>0$ there exists a unique $y>x$ such that $f(x) = g(y)$. Let $\tau: \mathbb{R}_{>0} \rightarrow \mathbb{C}$ be the curve $\tau: x \mapsto x+iy$, then $|\tau'(x)| \geq 1$ and $|\sinh(\tau)| = |\tau|$. So the curve

$$ x \mapsto \frac{\sinh(\tau)}{\tau} $$ maps into the unit circle. If we can bound its derivative from below, then it must wind around the unit circle (and therefore pass through $1$) infinitely many times. That this is indeed the case follows from the following inequalities:

$$ \left|\frac{\partial}{\partial x}\frac{\sinh(\tau)}{\tau}\right| \geq \left| \frac{\cosh(\tau)}{\tau} - \frac{\sinh(\tau)}{\tau^2} \right| \geq \left| \frac{\sqrt{\left| |\tau|^2-1 \right|}}{|\tau|} - \frac{1}{|\tau|} \right| \xrightarrow{x \rightarrow \infty} 1 $$

This shows that $\sinh(\tau) = \tau$ occurs infinitely many times. Moreover, all solutions of $\sinh(z) = z$ with $\Re(z)>0$ and $\Im(z) >0$ lie on $\tau$.

We can construct many solutions by setting $f(x)=F'(x)$ where $F$ is smooth and satisfies $$ 2F'(x)=F(x+1)-F(x-1)$$ for all $x$. For example, $F$ can be generated from this formula by starting with an arbitrary smooth ($C^\infty$) function on $[-1,1]$ that vanishes near the points $-1$, $0$, and $1$: For $x>1$, $$F(x)= F(x-2)+2F'(x-1)$$ defines $F$ successively on $(k,k+1]$ for $k=1,2,\dots$, and for $x<-1$, $$F(x)=F(x+2)-2F'(x+1)$$ does the same trick successively on $[-k-1,-k)$. This yields a function $F$ that vanishes near each integer, and is clearly smooth.