Let $f$ be a monotone nondecreasing function of a single variable which is also convex. Let $g$ be a convex function defined on a convex set $G$. Is it true that the composition of these functions $f \circ g$ is strictly convex if and only if $f$ is monotone increasing? Are there other requirements?
Further,
Let $f$ be twice continuously differentiable on a region $A \subset \mathbb{R}^n$. I am to show that a sufficient condition for a point $x^*$ in the interior of $A$ to be a relative minimum point of $f$ is that $f(x^*) = 0$ and that $f$ be locally convex at $x^*$. I am not sure how to go about showing this.
Thanks in advance!
One question per question, please.
Yes, because for $0\le \lambda\le 1$ and for every $a,b$ $$f(g((1-\lambda)a+\lambda b))\le f((1-\lambda)g(a)+\lambda g(b))\le (1-\lambda)f(g(a))+\lambda f(g(b))) $$
Suppose $x^*$ is not a point of relative minimum. There there is a nearby point $x^{**}$ with $f(x^{**})<f(x^*)$. Consider the function $g(t)=f((1-t)x^*+tx^{**})$ for $0\le t\le 1$. This function is convex (why?). Since $g(t)\le (1-t)f(x^*)+tf(x^{**})$, it follows that $$ \frac{g(t)-g(0)}{t} \le f(x^{**}) - f(x^*) <0 \tag{1}$$ On the other hand, $g'(0)=0$ by the chain rule. This contradicts (1).