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Let $f(n)$ be non-negative real valued function defined for each natural number $n$.

If $f$ is convex and $lim_{n\to\infty}f(n)$ exists as a finite number, then can we conclude that $f$ is non-increasing?

user74261
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2 Answers2

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This is true: Recall that $f$ is convex if and only if for every fixed $x$ the function $g(y)=\frac{f(x)-f(y)}{x-y}$ is non-decreasing. Now if $g(y_0)=c>0$ for some $y_0$ then for $y\geq \max\{ x,y_0\}$ we get that $f(x)-f(y)\leq c(x-y)$, or in other words, $f(y)\geq f(x) +c(y-x)$. Therefore, by letting $y\to \infty$, we see that $f$ can't have a finite limit. Therefore $g\leq 0$ and this is equivalent to $f$ being non-increasing.

Jose27
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If $f$ is not non-increasing, there exist $x_1$ and $x_2$ such that $x_1 < x_2$ and $f(x_1) < f(x_2)$. The secant through $(x_1,f(x_1))$ and $(x_2,f(x_2))$ has positive slope, and by convexity, the graph of $f$ lies above that secant on $(x_2,\infty)$. Therefore $\lim_{x\to\infty} f(x) = \infty$. Contraposing, if $\lim_{x\to\infty} f(x)$ exists and is finite, then $f$ is non-increasing.