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?


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 $x\in\mathbb R$.

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

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