Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve a problem which is as follows:

Suppose $f$ is a convex function on a radially open convex subset $C$ of a vector space $E$, and $x \in C$. Show that there exists a linear functional $l$ on $E$ such that $f(x+z) \geq f(x) + l(z)$ for all $z$ such that $x+z \in C$. Now I've not progressed very far on this problem, but I did the only obvious thing so far and rewrote the above as (fixing $z$ here) $l(z) \leq f(x+z) - f(x)$ for all $x$ where $x+z \in C$: denote this set of $x$ by $C(z,x)$ for brevity.

Then we can immediately deduce $l(z) \leq inf_{x \, \in \, C(z,x)}\,(f(x+z)-f(x))$, so $l$ is a sort of lower bound for $f(x+z)-f(x)$, when the right things are defined. Since we're working in such generality with all convex functions, all I could think of to do here was to define $l(z) := inf_{x \, \in \, C(z,x)}\,(f(x+z)-f(x))$ wherever $C(z,x)$ is not empty: of course if the infimum is not defined then $l$ cannot be defined so that hopefully won't present any issues, then perhaps we can use some sort of Extension Theorem to define $l$ on all of $E$. However, I'm not even convinced this is the right $l$ (it doesn't appear to be linear to me, for one thing), but I can't think of anything else to try which is general enough that we can define it for any convex function. Am I close, or could anyone help point me in the right direction if not? Many thanks.

(Hints or solution are both fine - I'm not handing this work into anyone though so hints will probably be more useful, but maybe a solution would help if I still can't come up with anything. The original question had multiple parts about directional derivatives if that's any use, so some of the information given (radially open, etc) may turn out to be superfluous here.)

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.