Functions $f:\mathbb Z^2\to\mathbb R$ with $f(x,y)=\frac{f(x-1,y)+f(x,y-1)}{2}$ Find all functions $f:\mathbb Z^2\to[0,1]$ such that for any $x,y\in\mathbb Z$, we have
$$f(x,y)=\frac{f(x-1,y)+f(x,y-1)}{2}.$$
Any constant function works, and I'm tempted to believe that they are the only solutions.
By inductively applying the relation, I have shown that
$$f(x+n,y+n)=2^{-n}\cdot\sum_{k=0}^n\binom nkf(x+k,y+n-k).$$
I have not really made more progress from here.
 A: Starting at $(x,y)$, take a random walk $S_n$ down and left. The condition implies that $M_n=f(S_n)$ is martingale. As $f$ is bounded, by the martingale convergence theorem, $M_n$ converges (say to $M_\infty$) almost surely.
So let $X_1,X_2\dots$ be the individual left/down steps of the random walk. By the Hewitt-Savage zero-one law, for any interval $[a,b]\subset\mathbb R$ we have
$$\mathbb P\left(\lim_{n\to\infty} f((x,y)+X_1+\dots+X_n)\in[a,b]\right)\in\{0,1\}.$$
Thus there exists $\lambda\in\mathbb R$ such that
$$\mathbb P\left(\lim_{n\to\infty} f((x,y)+X_1+\dots+X_n)=\lambda\right)=1.$$
As $f$ (and hence $M_n$) is bounded, it follows by the dominated convergence theorem that
$$f(x,y)=\mathbb E[M_0]=\lim_{n\to\infty}\mathbb E[M_n]=\mathbb E[M_\infty]=\lambda$$
Finally, note that for any $(x',y')\neq(x,y)$, the two random walks starting at $(x,y)$ and $(x',y')$ meet with positive probability, so we must also have $f(x',y')=\lambda$ too, i.e. $f$ is constant.

This is similar to the problem that a lower-bounded discrete harmonic function on $\mathbb Z^2$ is constant (see e.g. here). That problem could be solved by high-tech probabilistic arguments, which motivated me trying the same for this one.
The argument actually ended up being a little more fiddly for this problem.
