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$f:\mathbb{N}\to \mathbb{R}$ is bounded above and satisfies $$f(n)\le \frac{f(n+1)+f(n-1)}{2}$$ Does it follow $f$ is constant ?

I assumed $f$ achieves a maximum $M$ , suppose $n_0$ is the smallest solution of $f(x)=M$ then applying the conditions for $n=n_0$ we reach a contradiction, but since the range is $\mathbb{R}$ it may not achieve maximum, for example $f(n)=2-\frac{1}{n}$ is bounded above by $2$ but it is never reached, so we have a problem there. Any help will be welcome. Also, if we remove boundedness, we see any convex function works, so boundedness is necessary. Thanks in advance for all help.

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I think your proof of impossibility for achieving the maximum is flawed - why can the maximum not be attained for $n=1$ - and consider decreasing functions. – Mark Bennet Jun 16 '14 at 6:31
Hi there is a typo with the question, I am sorry everyone, I will edit it. the domain is not $\mathbb{N}$, its $\mathbb{Z}$. And yes my argument is wrong here. – shadow10 Jun 16 '14 at 6:34
@Joybangla Since there are two answers already, one (really both) of which depends quite heavily on the domain being $\mathbb{N}$, I think it would be better to just ask a new question with the correct domain. Link back to this one, and explain the difference. – user61527 Jun 16 '14 at 6:38 I have posted it here, sorry for all the troubles. Can someone help me now? – shadow10 Jun 16 '14 at 6:42

$f(x) = \frac{1}{x}$ is convex, so I think $f(n) = \frac{1}{n}$ will work (bounded above by 1). The maximum is achieved at $1$, so there is no $n-1$ there for the argument to work.

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I think that $$f(n)\le \frac{f(n+1)+f(n-1)}{2}$$ only means that $f(n)$ is convex. So your $f(n)$ is cup-shaped (or U-shape) and it reaches its maximum value(s) at the the end(s).


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