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I'm trying to solve this problem.

Let $ f $ be a convex function such that $ f: \mathbb R\rightarrow \mathbb R. $ Prove that $$ \lim_{x\rightarrow +{\infty}}f(x) $$ and $$\lim_{x\rightarrow -{\infty}} f(x) $$ always exists. If those limits are both finite, it is true that $ f$ is a constant function?

Thank you for the help!

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  • $\begingroup$ If it ever increases, ask the question: is it allowed to decrease again? $\endgroup$ – Michael May 31 at 19:55
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Since $f$ is convex, given $s < t < u$, we have $$ \frac{f(t) - f(s)}{t - s} \le \frac{f(u) - f(s)}{u - s} \le \frac{f(u) - f(t)}{u - t} $$

If $f$ is constant, the limits exist.

If not, suppose $f(s) < f(t)$. Letting $u \rightarrow + \infty$ in the first inequality, we get $\lim_{x \rightarrow +\infty} f(x) = +\infty$. Now, suppose $f(t) > f(u)$. Letting $s \rightarrow - \infty$ in the second inequality, we get $\lim_{x \rightarrow -\infty} f(x) = +\infty$.

Now, assume both limits exist and are finite. Letting $u \rightarrow + \infty$ in the first inequality, we get $f(t) \le f(s)$ and letting $s \rightarrow - \infty$ in the second inequality, we get $f(t) \le f(u)$. So $f$ is both non-increasing and non-decreasing, i.e., constant.

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For the first part, specifically, $\limsup_{x\rightarrow\infty} f(x)=\liminf_{x\rightarrow\infty}f(x)$.

To see this, let $s_n$ be the sequence of $x$ that achieves the sup and $m_n$ be the one for inf. Then

$$f(tm_n+(1-t)m_{n+1})\leq tf(m_n)+(1-t)f(m_{n+1}).$$

As both $m_n,s_n$ go to infinity, you can always find a subsequence $s_{n_k}$ which interlaces between infinitely many $m_{n_k},m_{n_{k+1}}$. It follows that $f(s_{n_k})\leq t_{n_k}f(m_{n_k})+(1-t_{n_k})f(m_{n_{k+1}})$. Take the limsup of both sides to conclude that $\limsup_{x\rightarrow\infty}f(x)\leq\liminf_{x\rightarrow\infty}f(x)$.

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