What is a reasonable finitary analogue of the statement that harmonic functions are smooth? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:20:11Z http://mathoverflow.net/feeds/question/33602 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a What is a reasonable finitary analogue of the statement that harmonic functions are smooth? Qiaochu Yuan 2010-07-28T02:43:26Z 2010-08-20T20:47:43Z <p>In my answer to <a href="http://math.stackexchange.com/questions/640/why-are-differentiable-complex-functions-infinitely-differentiable/820#820" rel="nofollow">this question</a> on MU, I suggested that the OP think about the difference between real-differentiable and complex-differentiable functions by using a sort of finitary analogue. One way to make this precise is the following. The finitary analogue of a real-differentiable function is just a function $f : \mathbb{Z} \to \mathbb{Z}$, whose finitary derivative is the first difference $f(z + 1) - f(z)$. The first difference can be arbitrarily nasty; in particular it does not have to grow in any kind of smooth way, which is a reasonable analogue of what happens for arbitrary real-differentiable functions.</p> <p>The finitary analogue of (the real part of) a complex-differentiable function, on the other hand, is a discrete harmonic function $f : \mathbb{Z}^2 \to \mathbb{Z}$ e.g. one which satisfies </p> <p>$$\frac{f(x+1, y) + f(x-1, y) + f(x, y+1) + f(x, y-1)}{4} = f(x, y).$$</p> <p>Such functions are clearly more constrained than arbitrary functions $\mathbb{Z} \to \mathbb{Z}$. Is there a reasonable finitary statement about the properties of such functions which is analogous to the statement that harmonic functions are smooth?</p> <p>(The best case would be if one could deduce the infinitary statement from the finitary one, but maybe this is difficult. Also, I can't think of good tags for this question, so feel free to retag.)</p> http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/33607#33607 Answer by Daniel Litt for What is a reasonable finitary analogue of the statement that harmonic functions are smooth? Daniel Litt 2010-07-28T03:29:23Z 2010-07-28T03:47:12Z <p>One reasonable thing to note is that defining $$f'(x, y)=f(x+1, y)+f(x-1, y)-f(x, y-1)-f(x, y+1)$$ we have that $f'$ is also harmonic in this sense. So we have that arbitrary "derivatives" of harmonic functions are harmonic, though this is much more trivial than in the smooth case. </p> <p>I don't see how to get the smooth case from this sort of remark, but I do think that this captures the notion of the derivative of a harmonic function to some extent, in that it measures some sort of rotation and scaling (as you point out in your answer to the math.SE question). </p> http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/33608#33608 Answer by Per Vognsen for What is a reasonable finitary analogue of the statement that harmonic functions are smooth? Per Vognsen 2010-07-28T03:37:59Z 2010-07-28T03:46:30Z <p>Did you first try working out what happens for one-variable discrete harmonic functions? You said that "such functions are clearly more constrained than their one-variable counterparts" but must have meant one-variable functions that are real-differentiable rather than harmonic. For one-variable discrete harmonic functions are <em>very</em> constrained:</p> <p>If you know $f(x+1)$ and $f(x-1)$ you can first interpolate $f(x)$ and then extrapolate $f(x+2)$ and $f(x-2)$. By induction, every value of $f$ is determined this way. That settles the special case of the standard suite of theorems (minimum-maximum principle, Liouville's theorem, mean-value theorem, etc) for a unit interval.</p> <p>You can piece together unit intervals to prove the more general case. That piecing-together logic is similar to the proof of Green's theorem that reticulates the domain and passes to the limit, so you could say this extends to the infinitary case, modulo analytic due diligence.</p> <p>As for smoothness (harmonic functions are actually real-analytic), I'm not sure of the proper discrete analogue, so I can't help you there.</p> http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/33610#33610 Answer by T. for What is a reasonable finitary analogue of the statement that harmonic functions are smooth? T. 2010-07-28T03:44:39Z 2010-07-28T06:29:02Z <p>There are quite a few properties shared by discrete and continuous harmonic functions. They also generalize in various forms to graphs. I don't know if the analogy is so close that there is a unique concept of smoothness for the discrete case, but see the last item for one point of contact.</p> <ol> <li><p>Poisson formula: $f(p)$ is a weighted average of the values of $f$ on a "curve" of points surrounding $p$ (that is, a minimal set of points disconnecting it from infinity). As in the continuous case the weighting is the distribution of hitting probabilities, $h_p(x,y)$ = probability that a random walk started at $p$ first hits the curve at $(x,y)$.</p></li> <li><p>Maximum principle. Because values in the interior are weighted averages of values on the boundary. </p></li> <li><p>Growth constraints. A bounded discrete-harmonic function is constant. More generally there is a discrete analogue of the theorem that if a harmonic $f(x,y)$ has polynomial-bounded growth, $|f(x,y)| &lt; C(|x|+|y|+1)^n$ then $f(x,y)$ is a (harmonic) polynomial in $x$ and $y$, of degree at most $n$. </p></li> <li><p>Harnack inequalities. It is these that are a form of smoothness. They say that if $a$ and $b$ are nearby points far away from the boundary curve used in the Poisson formula, $f(a) - f(b)$ is small (relative to $|f|$ on the boundary). There are many expressions of this principle, see for example papers of Fan Chung and her collaborators ( www.math.ucsd.edu/~fan ) for inequalities in the case of graphs with a vertex-transitive symmetry group. By phrasing the theory in terms of eigenfunctions of the Laplacian one can give the discrete and continuous Harnack inequalities a very parallel form. But in the discrete case they also seem to play a more basic role of showing that smoothness is present in some form, which in the continuous case is automatic because $f$ is assumed to have at least two derivatives. </p></li> </ol> http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/33644#33644 Answer by Spencer for What is a reasonable finitary analogue of the statement that harmonic functions are smooth? Spencer 2010-07-28T11:25:09Z 2010-07-28T11:25:09Z <p>There are a few tidbits in <a href="http://terrytao.wordpress.com/2010/02/18/a-proof-of-gromovs-theorem/" rel="nofollow">this</a> blog post of Tao's from a few months back. He talks about Lipschitz harmonic functions on groups (it's in the context of Gromov's theorem about finitely generated groups of polynomial growth). There is mention of 'maximum principles' and elliptic regularity but having not been through the details myself I can't say that I am confident with the analogy.</p>