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Problem: Imagine an infinite chessboard that contains a positive integer in each square. If the value of each square is equal to the average of its four neighbors to the north, south, west, and east, prove that the values in all the squares are equal.

I understood the problem can be proved by using the decreasing sequence. I think it is false when 'positive integer' changed to 'positive real number', but I can't prove it.

Question: Is it still true when 'positive integer' changed to 'positive real number'?

Thank you for your advice.

Yi Master
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  • Hi, genuinely curious here. Would you mind illustrating how you solved the case with positive integers? – K.defaoite Jun 03 '20 at 14:40
  • Choose some square in chessboard(called 'first square'). If "The values in all the squares are equal" is false, there exists a square around the first square such that a number in square is smaller than a number in the first square. Using this property, we can make the decreasing sequence. – Yi Master Jun 03 '20 at 15:00
  • Ah, I see. I think then that you are correct - this is not true for positive real numbers. But I think this is tricky to prove. The only way I see of doing this is actually constructing a solution which could be very difficult. – K.defaoite Jun 03 '20 at 15:08
  • Thanks so much. – Yi Master Jun 03 '20 at 15:15
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    Answered here. Generally the assumption strongly resembles the definition of a harmonic function, which by a theorem of Liouville must be constant if bounded from below. It seems that the same is true in the discrete case. – Adayah Jun 03 '20 at 17:30

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Intuitively speaking:

W.L.O.G let, $f(0,0)=0$. There is no minimum at the lattice points.

And let, for $x,y \to -\infty$ $f(x,y)$ converges to $M$.

So, for some $x,y<<0$, let, $|f(x,y)-M|<\epsilon$, for some $\epsilon >0$.

The 4-tuple of differences between the points and it's neighbours will be like $(-\epsilon{_1}, -\epsilon'{_1}', -\epsilon{_2}, \epsilon'{_2})$.

Such that $-\epsilon{_1}-\epsilon'{_1}+\epsilon{_2}+\epsilon'{_2}=0$.

And, if you start extending the structure from here. You will never be able to maintain $|f(x,y)-M|<\epsilon$ in its neighbourhood. If you try to maintain this convergence requirement in one portion , you fail to do in a opposite portion.

So, the convergence must not be true. It must go to $-\infty$. This means $f$ can be bounded neither from above nor from below with being it a constant function. So, $f(x,y)$ must have to be a constant function.

Alapan Das
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