2
$\begingroup$

Is there some strictly convex function defined in $\mathbb{R}$ to be unbounded(above and lower)? For example, $f:(\infty,0]\to \mathbb{R},$ $f(x)= -x^2$ is a strictly convex function. However, this function is not defined in all $\mathbb{R}$ and is bounded above.

Other case:$f:\mathbb{R}\to \mathbb{R},$ $f(x)= x$ is a convex function wit all requirements but is no a strictly convex function.

Some ideias?

$\endgroup$
6
$\begingroup$

I would say that $f(x)=-x^2$ is anyway defined on all $\mathbb R$. Moreover, I think that the usual definition of a convex function (see https://en.wikipedia.org/wiki/Convex_function) is the opposite with respect to the one you adopted here, so $f(x)=x^2$ is indeed convex.

Having said that, you can by sure find a strictly convex function $f(x)$ defined on the whole real line such that $$ \lim_{x\to\pm\infty}f(x)=\pm\infty. $$ For example, $$ f(x)=x+e^x $$ is such a function, since $f''(x)=e^x>0$ for any $x\in\mathbb R$.

$\endgroup$
  • $\begingroup$ Yes. You are right. $f(x) = - x^2$ is concave. $\endgroup$ – orrillo Jun 12 '16 at 22:09

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.