# Convex functions - two questions

I have two questions regarding convex functions:

First question: Let f be convex function on closed interval [a,b].

Prove that f has maximum in x=a or x=b.

I understand that $\forall x\in[a,b]: \ \ f''(x)>0$, thus $f'(x)$ is monotonically increasing and the slope of the tangent is increasing. Is there any way to continue from here? I also tried to use the definition of convex function, but it didn't help...

Second question: Let $f:(0,\infty)\to\mathbb{R}$ be convex function and $\lim_{x\to0^{+}}f(x)=0$.

Prove that $g(x)=\frac{f(x)}{x}$ is monotonically increasing in $(o,\infty)$.

Eventually, I solved the second question. Still looking for help with the first one.

Please help, thank you!

• I edited my question. May you help? – Galc127 Jan 27 '14 at 20:28

## 3 Answers

Suppose that neither $a$ nor $b$ is a local maximum. Then, there exist $x,y\in(a,b)$ (not necessarily different) such that $f(x)>f(a)$ and $f(y)>f(b)$. This is easily shown to imply that $\max\{f(x),f(y)\}>\max\{f(a),f(b)\}$. Now, since $x,y\in(a,b)$, there exist $\lambda,\mu\in(0,1)$ such that $x=\lambda a+(1-\lambda)b$ and $y=\mu a+(1-\mu)b$. The convexity of $f$ implies that \begin{align*} f(x)\leq&\,\lambda f(a)+(1-\lambda)f(b)\leq\lambda\max\{f(a),f(b)\}+(1-\lambda)\max\{f(a),f(b)\}=\max\{f(a),f(b)\},\\ f(y)\leq&\,\mu f(a)+(1-\mu)f(b)\leq\max\{f(a),f(b)\}. \end{align*} This, in turn, implies that $\max\{f(x),f(y)\}\leq\max\{f(a),f(b)\}$, which contradicts our earlier result.

Main idea:

Suppose that $f$ attains its maximum at some $x_0\in(a,b)$. The main idea is that the point $[x_0,f(x_0)]\in\text{graph}(f)$ lies under the line segment connecting $[a,f(a)]$ and $[b,f(b)]$ (or, in the worst case, on this line segment). One can easily see that this can happen only in the case of $f(a)=f(x)=f(b)$.

If you define local maximum by having a neighbourhood around a or b where the function values are strictly lower than f(a) resp. f(b) then your edited question also makes no sense.

• Ok sorry, local maximum seems to be commonly defined by $f(a)>= f(x)$ – Stefan Bubble Jan 27 '14 at 20:37