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$f:\mathbb{Z}\to \mathbb{R}$ is bounded above and satisfies $$f(n)\le \frac{f(n+1)+f(n-1)}{2}$$ Does it follow $f$ is constant ? There was a dreadful typo in the previous question (in the previous question, the domain of $f$ was $\mathbb N$, here, it is $\mathbb Z$), I am posting a new one. Thanks for helping.

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shadow10
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    Look at $g(n)=f(n)-f(n-1)$ - what can you say about $g$? – Mark Bennet Jun 16 '14 at 06:45
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    @5xum Someone edited the previous question to be identical to this one. – Mark Bennet Jun 16 '14 at 06:46
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    @StevenTaschuk I retracted my vote. I will also edit this question so no-one will make the same mistake as me. – 5xum Jun 16 '14 at 06:48
  • To all of you sirs, I am terribly sorry I made a mistake in writing down the problem and sorry you all had to go through that trouble. Thanks a lot for all helpful comments. – shadow10 Jun 16 '14 at 06:54

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Here is an extended hint

Let $g(n)=f(n)-f(n-1)$, or $f(n)=f(n-1)+g(n)$ so the condition becomes $$g(n+1)\ge g(n)$$

Suppose $g(N)=a\gt 0$, can you say anything about $f(N+r)$?

Suppose $g(N)=b\lt 0$, what can you say about $f(N-r)$?

Mark Bennet
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