Find all continuous and bounded functions $g$ with : $$\forall x \in \mathbb R, 4g(x)=g(x+1)+g(x-1)+g(x+\pi)+g(x-\pi).$$

I have posted this question here, but received no answer.

  • 5
    $\begingroup$ MO is the wrong place to repost unanswered questions of this nature $\endgroup$
    – FShrike
    Feb 5 at 12:28
  • $\begingroup$ I have try in others forums, but no body have an idea $\endgroup$
    – Dattier
    Feb 5 at 12:31
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    $\begingroup$ "Very difficult" phrase makes me smile and quitting reading the rest of the text right away. $\endgroup$
    – Wlod AA
    Feb 5 at 12:33
  • 1
    $\begingroup$ If view it in a two dimension space, then the structure is discrete harmonic function, but it is bounded so it is constant. $\endgroup$
    – katago
    Feb 5 at 12:34
  • 1
    $\begingroup$ @WlodAA, re, I agree; such editorialising does not belong in MO titles. I edited it out. $\endgroup$
    – LSpice
    Feb 6 at 15:06

4 Answers 4


$\newcommand\de\delta$Considering $g$ a distribution (in the generalized-function sense), let $\hat g$ be the Fourier transform of $g$. Then your functional equation yields $$4\hat g(t)=e^{it}\hat g(t)+e^{-it}\hat g(t)+e^{i\pi t}\hat g(t)+e^{-i\pi t}\hat g(t),$$ or $$(\cos t+\cos\pi t-2)\hat g(t)=0,$$ for real $t$.

The equality $\cos t+\cos\pi t-2=0$ for real $t$ implies $\cos t=1=\cos\pi t$ and hence $t=0$ (because $\pi$ is irrational). So, the support of $\hat g$ is $\{0\}$. So (see e.g. "For every compact subset $K\subseteq U$ there exist constants $C_{K}>0$ and $N_{K}\in \mathbb {N}$ such that for all $f\in C_{c}^{\infty }(U)$ with support contained in $K$ [...]" here), we have $\hat g=\sum_{j=0}^n c_j\de^{(j)}$ for some $n\in\{0,1,\dots\}$ and some complex $c_j$'s, where $\de^{(j)}$ is the $j$th derivative of the delta function. So, $g$ is a polynomial. Since $g$ is bounded, it is constant. $\quad\Box$

  • $\begingroup$ This idea also looks like it can be used to prove the mean value formula of harmonic function. $\endgroup$
    – katago
    Feb 5 at 14:13
  • $\begingroup$ @katago : Sorry, I don't what "the mean value formula of harmonic function" is. $\endgroup$ Feb 5 at 14:19
  • $\begingroup$ Sorry, I mean, using your method seems we can prove $u$ satisfied the mean value property $\Longrightarrow$ $u$ is a harmonic function. If $$ u(x)=\frac{1}{\left|\partial B_r\right|} \int_{\partial B_r(x)} u d S=\frac{1}{\left|B_r\right|} \int_{B_r(x)} u(y) d y $$, let $\hat{u}$ be the Fourier transform of $u$, then $\hat{u}(x)=\frac{1}{\left|\partial B_r\right|} \int_{\partial B_r(x)} \hat{u}(x)e^{iv} d S=\frac{1}{\left|B_r\right|} \int_{B_r(x)} \hat{u}(x)e^{iv} d y$, then $\hat u(x)(1-\int_{\partial B_r(x)}vdS)=0$. but I do not know how to move on. $\endgroup$
    – katago
    Feb 5 at 16:00

(Compare On equation $f(z+1)-f(z)=f'(z)$.)

Plug $g(x)=e^{\lambda x}$. We obtain the characteristic equation: $$4=2\cosh \lambda+2\cosh\pi\lambda.$$ This has one real root $\lambda=0$ of multiplicity $2$. This root gives a solution $g(x)=ax+b$ with arbitrary constants $a$ and $b$. Since you are looking for bounded solutions, one has to set $a=0$ and we recover the constant solution of @katago. However there are infinitely many others: Let $\lambda=p+iq$ be any complex solution (there are infinitely many of them). Then any linear combination $$g(x)=\sum c_ke^{\lambda_k x}$$ gives you a solution. In general, this solution is complex, but if you want a real solution, notice that $\lambda_k$ come in complex conjugate pairs, so you have an infinte-dimensional space of real solutions. Now solutions bounded on the real line will correspond to pure imaginary $\lambda$, but the characteristic equation does not have pure imaginary roots.

So every bounded solution is indeed constant. That all solutions are linear combinations of exponentials or their limits is a general theorem cited in the reference that I gave in the beginning.


It seems $g$ can only be a constant function. First, $g$ is constant in any shift of set $f(A_x)=\{x+a+b\pi| a,b\in \mathbb N\}$ by Liouville type theorem for discrete harmonic function, and then we need suitable choose the constant to make the function continuous on $\mathbb R/A_x$. but by Dirichlet approximation theorem we can get $g$ is constant on $\mathbb R$.

  • $\begingroup$ If change the condition to sublinear instead of bounded then, the function may not be constant. $\endgroup$
    – katago
    Feb 5 at 13:22

Iosef's answer shows the only solutions are constants.

Plug …

Edgar, G. A.; Rosenblatt, J. M., Difference equations over locally compact Abelian groups, Trans. Am. Math. Soc. 253, 273-289 (1979). ZBL0417.43006.

At the end, we construct a nontrivial, continuous, bounded, almost-periodic function $F : \mathbb R \to \mathbb C$ satisfying $$ 0 = F(x+1)- F(x-1) -\sqrt{2}F(x+2\pi)+\sqrt{2}F(x-2\pi) . $$


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