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Let $f$ be a function that maps $\mathbb{Z}^2$ to $\mathbb{R}$ and consider the operator $T$ which replaces the value of $f$ at $(i,j)$ by the average of the values of $f$ at its four neighbors: $$ Tf(i,j) = \frac{f(i-1,j) + f(i+1,j) + f(i,j-1) + f(i,j+1)}{4}.$$ Now we know that the equation $$ Tf = f$$ does not have any solutions $f$ that are bounded. I'm looking for a stronger version of this fact. In particular, I want to know if there is a quantitative statement to the effect that if $$\inf_{(i,j) \in \mathbb{Z}^2} f(i,j) = 0 , ~~~~~~\sup_{(i,j) \in \mathbb{Z}^2} f(i,j) = 1, ~~~~~~~~~~~~~~~(*)$$ then $Tf$ is in some sense far from $f$?

One naive way to make this work would be to define $$ d_{\infty}(f,g) = \sup_{(i,j) \in \mathbb{Z}^2} |f(i,j) - g(i,j)| $$ and consider $$ \inf_f d_{\infty}(f,Tf) $$ where the infimum is taken over all $f$ satisfying $(*)$. But it isn't hard to see that this infimum is $0$, so this attempt fails. Perhaps a more sophisticated notion of distance would make this sort of statement true?

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    what is $x_i$. Moreover all norms in $\mathbb{R}^2$ are equivalent so what do you mean by more sophisticated notion of distance?2012-06-05
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    I rewrote the question is a more explicit and clearer way.2012-06-06
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    It seems that $d_1(f,g)=\sum_{(i,j)\in\mathbb Z^2}|f(i,j)-g(i,j)|$ would work, in the sense that $d_1(f,Tf)\ge c>0$ for all $f$ with normalization (*). "Proof by example": if $f(i,j)=(1-\epsilon\max(|i|,|j|))^+$, then $|f-Tf|$ is of size about $\epsilon$ on the boundary (and diagonals) of a square of size $1/\epsilon$. So the $\ell_1$ norm appears to be the right thing to use here. I don't have a proof, but could try if you are interested in this kind of result.2012-06-09

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