# Harmonic Functions

Suppose $$f: \mathbb{R}\times\mathbb{R}\to\mathbb{R}$$ is has continuous partial derivatives and

$$4f(x,y)=f(x+\delta,y+\delta)+f(x-\delta,y+\delta)+f(x-\delta,y-\delta) + f(x+\delta,y-\delta)$$

for all $$(x,y)$$ in $$\mathbb{R}\times\mathbb{R}$$ and all $$\delta$$ in $$\mathbb{R}$$.

I don't believe that $$f$$ is necessarly harmonic but I cannot construct a counter-example. Is $$f$$ harmonic?

• The easier step is to assume that $f$ is $C^2,$ that is continuous second partials, maybe $C^3$ if that helps. Then write the finite Taylor expansion around any point and see what that tells you. But your property does resemble the mean value (in)equalities, see Gilbarg and Trudinger, Elliptic Partial Differential Equations of Second Order. – Will Jagy Jun 9 '10 at 18:14

Yes, the Taylor series works. Actually $C^2$ suffices for the remainder term, although my sophomore calculus book gives the proof using $C^3.$ I get $$4 f(x_0, y_0) = 4 f(x_0, y_0) + \left( 2 f_{xx}(x_0, y_0) + 2 f_{yy}(x_0, y_0) \right) \delta^2 \; + \; o( \delta^2 )$$ and $$\left( 2 f_{xx}(x_0, y_0) + 2 f_{yy}(x_0, y_0) \right) \delta^2 \; = \; o( \delta^2 )$$ and $$2 \left( f_{xx}(x_0, y_0) + f_{yy}(x_0, y_0) \right) \; = \; 0$$

LATER EDIT: unless I am vastly mistaken this argument still works if we put in the caveat $| \delta | < \Delta = \Delta(x_0, y_0),$ that is we only require your equation for small $\delta$ and even say that the allowable size of $\delta$ depends on the position of the center point that I am calling $(x_0, y_0).$ But with this change we can build an easy discontinuous example of your relation, take $$f(x_0, y_0) = 1, \; \; if \; \; y_0 > 0,$$ $$f(x_0, y_0) = 0, \; \; if \; \; y_0 = 0,$$ $$f(x_0, y_0) = -1, \; \; if \; \; y_0 < 0.$$

Then your relation holds for $| \delta | < | y_0 |$ when $y_0 \neq 0$ and holds for all $\delta$ when $y_0 = 0.$

• It is probably to late in the year for it to be homework. – Steven Gubkin Jun 9 '10 at 18:54
• It would be interesting to know if there is a discontinuous function $f$ which satisfies the discrete Laplace equation at every point and for every $\delta>0$. – Andrey Rekalo Jun 10 '10 at 2:58
• Hi, Andrey. I had the feeling that such an example might be one of those axiom of choice things, but really elaborate even then. I certainly do not know how to construct it. All I can think of is constructing a non-measurable function by taking, on the real line, one representative for each equivalence class of the relation $x \sim y$ if and only if $x−y$ is rational. So wait, that trick shows that we can make a badly discontinuous example that works for all RATIONAL $\delta > 0.$ Points are equivalent if both $x$ and $y$ coordinates differ by rationals. Function constant on each class. – Will Jagy Jun 10 '10 at 3:51

$f$ is harmonic under the weaker assumption that it is just continuous.

Multiplying the identity $$f(x+\delta,y+\delta)+f(x-\delta,y+\delta)+u(x-\delta,y-\delta) + f(x+\delta,y-\delta)-4f(x,y)=0$$ with a test function $g\in C_0^{2}(\mathbb R^2)$ and integrating the result over $\mathbb R^2$, it's easy to see that $$\int_{\mathbb R^2}\left(g(x+\delta,y+\delta)+g(x-\delta,y+\delta)+g(x-\delta,y-\delta) + g(x+\delta,y-\delta)-4g(x,y)\right)f(x,y)\ dxdy=0.$$ Applying to the latter equality the argument in Will's answer, we obtain that, for every $g\in C_0^{2}(\mathbb R^2)$ $$\int_{\mathbb R^2}(g_{xx}+g_{yy})f\ dxdy=0. \qquad\qquad (*)$$

By a theorem of Kellog, every continuous solution $f$ to $(*)$ satisfies the mean value property $$f(P)=\frac{1}{2\pi}\int_{0}^{2\pi}f(P+re^{i\phi})d\phi$$ for any $P\in \mathbb R^2$ and any $r>0$. Therefore, $f$ is harmonic.

Update. As Pietro Majer and BS indicated, the argument works for locally integrable $f$. This and the comment of Will Jagy above make me think that any discontinuous function that solves the discrete Laplace equation for every $\delta>0$ should have highly pathological properties. Perhaps, non-measurable solutions to Cauchy's functional equation $$f(x+y)=f(x)+f(y)$$ might be a close analogy.

• Will, I'm not sure the proof will work if the upper bound does not remain $>\epsilon>0$ for every $(x,y)$ in the support of $g$. The question is whether the passage to the limit $\delta\to 0$ in the integral can be justified in this case? – Andrey Rekalo Jun 10 '10 at 1:35
• Will, if we can choose however small but the same $\delta>0$ for every $(x_0,y_0)\in {\rm supp} g$, then we may write the Taylor expansion and safely pass to the limit in the integral. In your example, the upper bound on $\delta$ goes to $0$ when $y_0\to 0$. And the limiting function is discontinuous precisely on the line $y_0=0$. – Andrey Rekalo Jun 10 '10 at 1:59
• Excellent. Similar to the distinction between, say, Lipschitz and locally Lipschitz properties. – Will Jagy Jun 10 '10 at 2:16
• Yes, it's a good analogy. – Andrey Rekalo Jun 10 '10 at 2:49
• Your argument shows the result holds for any distributional $f$ (for instance a locally integrable $f$). – BS. Jun 10 '10 at 14:47

To reduce the case of a continuous $f:\mathbb{R}^2\to\mathbb{R}$ to, say the $C^2$ or $C^3$ case, one can simply mollify $f$ via convolution with a smooth kernel with compact support. Then $f_\epsilon:=f*\phi_\epsilon$ still enjoys the "N-S-W-E mean property" above, so it's harmonic as already seen in previous answers, and since as $\epsilon\to0$ the $f_\epsilon$ converge to $f$ uniformly on compact sets together with all second order derivatives, $f$ is harmonic too. Note that the analogous holds for a continuous $f$ (or even just locally integrable, it works as well) on an open subset of $\mathbb{R}^n$.

Variation: take $\phi$ as above and moreover with radial symmetry. Then $f*\phi$ is harmonic as before, so $f*\phi*\phi=f*\phi$ because harmonic functions are invariant by convolution with radial symmetric kernels (it's just a weighted radially symmetric mean value property). Since $\phi$ has compact support we can simplify $\phi$ in the last equality (this is standard via Fourier transform) and get $f*\phi=f$ so f itself is harmonic.