# How may I find all continuous and bounded functions g with the following property?

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

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

$$\newcommand\de\delta$$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. $$\quad\Box$$

• This idea also looks like it can be used to prove the mean value formula of harmonic function. Feb 5 at 14:13
• @katago : Sorry, I don't what "the mean value formula of harmonic function" is. Feb 5 at 14:19
• 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. 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 $$\mathbb R$$.

• If change the condition to sublinear instead of bounded then, the function may not be constant. 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 $$0 = F(x+1)- F(x-1) -\sqrt{2}F(x+2\pi)+\sqrt{2}F(x-2\pi) .$$