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I will focus on the real line. Let $f$ be a smooth function on $\mathbb{R}$, if $\forall x\in\mathbb{R}, r>0$, $$\frac{f(x-r)+f(x+r)}{2}=f(x),$$ we say that f has the spherical mean value property (MVP); if instead $$\frac{1}{2r}\int_{x-r}^{x+r}f(t)dt=f(x),$$ we say f has the ball MVP. It is easy to see that spherical MVP implies ball MVP.

It is well known, and not hard to prove (though it is much harder in higher dimensions), that ball MVP implies that $f$ is harmonic, i.e. $f$ is of the form $ax+b$, where $a$ and $b$ are constants.

Notice that in the definitions above, we require the radius $r$ to run over all the positive numbers. Out of curiosity, I tried to find non harmonic functions which satisfy the MVPs only for $r=1$. It turns out any $1$-periodic function will satisfy the spherical MVP with $r=1$, which is obvious. But it seems much harder to find one for ball MVP. So here is my question:

Question: Does there exist a function $f$ which is not of the form $ax+b$, such that $\forall x\in\mathbb{R},$ $$\frac{1}{2}\int_{x-1}^{x+1}f(t)dt=f(x)?$$

Thank you!

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    Just a side comment: for the mean value property to imply harmonicity, you don't need $r$ to run over all the positive numbers. It suffices (for example) that the property holds for all $r\in [0,\epsilon)$ for some $\epsilon > 0$.2012-03-05
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    @WillieWong: does $\epsilon$ depend on $x$?2012-04-13
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    it may. For $C^2$ functions it follows directly from the proof for MVP to imply harmonicity.2012-04-17
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    @WillieWong: Can we weaken the $C^2$ assumption into, say $C^0$? In practice $C^2$ seems hard to verify.2012-04-28
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    See page 17 of _Harmonic Function Theory_ by [Axler, Bourdon, and Ramey](http://axler.net/)2012-04-30
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    @WillieWong: Thanks!2012-04-30

2 Answers 2

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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 $\tau$.

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    Are you sure that $\sinh(z) = z$ has a nonzero solution?2012-03-04
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    @NickStrehlke Using Mathematica's function `FindRoot`you find many solutions like $z=13.9+3.35221\,I$ and $z=7.49768 + 2.76868$. I have not tried to prove that they are in fact solutions.2012-03-04
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    @WimC: Nice! Could you show why $\sinh(z)=z$ has a nonzero solution?2012-03-04
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    @SyangChen, just did (I hope). The proof is fairly contrived and I suspect Rouché can be applied in some clever way instead.2012-03-04
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    @NickStrehlke, looks like you swapped real and imaginary parts in (at least) your first root? Both $3.35221+13.9 i$ and $2.7868+7.4968 i$ appear to be close to a root of $\sinh(z) - z$.2012-03-04
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    @NickStrehlke: oops, I highlighted the wrong person. Should be Julián Aguirre, sorry.2012-03-04
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    @WimC Nice solution!2012-03-04
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    @NickStrehlke Yes, I posted solutions to $\sin a=a$. For such $a$, $\sin(a x)$ and $\cos(a x)$ are a solutions to the problem. I was going to write the answer, but I WinC had already done so.2012-03-04
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    @JuliánAguirre Well thanks anyways for mentioning Mathematica's FindRoot function, because I didn't know about that. I'm glad Syang Chen asked for a proof anyways, because that is also nice to see.2012-03-04
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    @WimC Even nicer proof for the existence of roots!!2012-03-04
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    @SyangChen I added an even shorter proof based on a theorem of Picard.2012-03-05
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    @WimC Awesome! I also thought about Picard's great theorem. But I had no idea how to exclude the possibility of $0$ being the exceptional point. Btw, is it possible to figure out the exceptional point (if it exists)?2012-03-05
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    @SyangChen There can be none. Same argument really: if some $w$ would occur only a finite number of times then so would $w + 2\pi i$.2012-03-05
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    @WimC Got it. So $f(z) - z$ is always surjective if $f$ is a periodic entire function. In particular periodic entire function always has a fixed point.2012-03-06
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    @SyangChen Correct. Note that "great Picard" is quite heavy machinery. Some statements only require "little Picard", for example to show that $f(z)-z$ must have at least one root. However, for the application at hand that wasn't sufficient.2012-03-06
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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.

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    Keen observation! Are there similar arguments in higher dimensions? Please see this MO post. http://mathoverflow.net/questions/90233/mean-value-property-with-fixed-radius2012-03-05