1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ I edited my question. May you help? $\endgroup$ – Galc127 Jan 27 '14 at 20:28
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $f(a)=f(x)=f(b)$.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.