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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 $\epsilon > 0$. –  Willie Wong Mar 5 '12 at 9:34
    
@WillieWong: does $\epsilon$ depend on $x$? –  Syang Chen Apr 13 '12 at 5:16
    
it may. For $C^2$ functions it follows directly from the proof for MVP to imply harmonicity. –  Willie Wong Apr 17 '12 at 4:30
    
@WillieWong: Can we weaken the $C^2$ assumption into, say $C^0$? In practice $C^2$ seems hard to verify. –  Syang Chen Apr 28 '12 at 3:54
    
See page 17 of Harmonic Function Theory by Axler, Bourdon, and Ramey –  Willie Wong Apr 30 '12 at 8:29
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2 Answers

up vote 8 down vote accepted

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? –  Nick Strehlke Mar 4 '12 at 10:49
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@NickStrehlke Using Mathematica's function FindRootyou 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. –  Julián Aguirre Mar 4 '12 at 11:54
    
@WimC: Nice! Could you show why $\sinh(z)=z$ has a nonzero solution? –  Syang Chen Mar 4 '12 at 16:27
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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. –  Julián Aguirre Mar 4 '12 at 20:05
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@WimC Even nicer proof for the existence of roots!! –  Syang Chen Mar 4 '12 at 22:52
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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. mathoverflow.net/questions/90233/… –  Syang Chen Mar 5 '12 at 1:09
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