Consider a non-constant harmonic function $f$ on $\mathbb{Z}^d$ (meaning this that $f(x)$ if the average of the $2d$ values $f(y)$ such that the distance between $x$ and $y$ is one). Let $M_n$ denote the maximum of the absolute values $|f(x)|$ for all $x$ such that $||x||_{\mathbf{L}^1} \leq n$ (where $n$ is a positive whole number). I am guessing that there is a constant $k>0$ such that, no matter what the whole positive number $n$ might be, $M_n \geq kn$. However, I am not sure how to tackle this. Any suggestions?
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1$\begingroup$ I'm not an expert in analysis, but I think this kind of argument works: Suppose $f$ is bounded by a polynomial. The Fourier transform $\hat{f} : T^n \to \mathbb{R}$ is then a tempered distribution. From the harmonic condition, we get $\hat{f} \cdot (\sum_k (e^{2 \pi i x_k} + e^{- 2 \pi i x_k}) - 2d) = 0$ and so $\hat{f}$ is supported on the origin. This implies that $\hat{f}$ is a linear combination of $D^\alpha \delta$ and thus $f$ is indeed a polynomial. $\endgroup$– Dongryul KimCommented Dec 8, 2016 at 2:54
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$\begingroup$ @DongryulKim A question concerning your last statement has appeared here just recently, mathoverflow.net/q/256523/41291 $\endgroup$– მამუკა ჯიბლაძეCommented Dec 8, 2016 at 5:55
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The way your statement is formulated, it is wrong. The simplest counterexample is the function $f(x,y)=xy$ on $\mathbb Z^2$ which vanishes on the ball of radius 1. In Heilbronn's pioneeriing 1949 paper it is, in particular, proved that any polynomial growth harmonic function for the simple random walk on $\mathbb Z^d$ is polynomial. This result is actually valid for any group of polynomial growth and any finitely supported averaging probability measure, see Alexopoulos https://projecteuclid.org/download/pdf_1/euclid.aop/1023481007.
However, the answer to the corrected version of your question, whether $$ \liminf \frac{M_n}n >0 $$ for any non-constant harmonic function on $\mathbb Z^d$, is "yes". Indeed, if the limit above is 0, then by the maximum principle applied to the $\ell^1$ balls in $\mathbb Z^d$, the growth of $f$ is at most linear, so that by the above theorem of Heilbronn, $f$ is actually linear, which yields a contradiction.
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$\begingroup$ What about saying that "$\dfrac{M_n}{n}$ is bound away from zero for $n$ large enough" $\endgroup$ Commented Dec 8, 2016 at 17:46
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$\begingroup$ Thank you so much! The paper of Heilbronn is what I needed. $\endgroup$ Commented Dec 13, 2016 at 0:30