convex and unbounded implies increasing?

Question

Suppose $f(x)$ is a positive continuous function on $(-\infty,\infty)$, symmetric about $0$. Let $f(x)$ is convex and $\lim_{x \rightarrow \infty} f(x) = \infty$. Can we say that $f(x)$ has to be increasing? I was trying to use the Taylor's theorem upto second order term but nothing is mentioned regarding the differentiability of $f(x)$. Can anyone help?

Answer 1

Yes, $f$ is necessarily increasing on $[0,\infty)$. Let $0\leq x<y$. By assumption we have $f(-y)=f(y)$ and hence by convexity: $$f(x)=f(\lambda(-y)+(1-\lambda) y)\leq \lambda f(-y)+(1-\lambda)f(y)=f(y)$$ For $\lambda=\frac{y-x}{2y}$.
Note that we didn't need the assumption $\lim_{x\to\infty}f(x)=\infty$. In fact if we just assume $f$ to be non-constant, it will already follow that $\lim_{x\to\infty}f(x)=\infty$