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I have a continously differentiable function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ which I am trying to prove is globally convex. Computing the Hessian directly is very difficult as it is a somewhat complicated function of a matrix, other methods of proving global convexity have proved inconclusive. So far I am only able to show that it is 'locally convex' in the following sense:

For any $x\in\mathbb{R}^{n}$ there exists an $\varepsilon_{x}>0$ such that for $y\in\mathbb{R}^{n}$ where $\| y-x\|\leq\varepsilon_x$ it holds that $$f(y)\geq f(x)+\nabla f(x)^{T}(y-x). $$

My question is a rather basic one, can we establish that local convexity of this kind implies global convexity? Are any extra conditions needed?

My intuition suggests that a continuously differentiable function on a convex set which is locally convex everywhere should be globally convex, but I have trouble constructing the argument. Any help is greatly appreciated!

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  • $\begingroup$ If $f$ is twice differentiable, your local condition imply that the hessian is positive semidefinite, which again imply global convexity. $\endgroup$
    – user251257
    Jun 2 '20 at 21:30
  • $\begingroup$ Thanks for your comment. Can you give me a hint toward a proof, perhaps via Taylor expansion? $\endgroup$
    – slyyah
    Jun 2 '20 at 21:34
  • $\begingroup$ Is the function $C^2$? If so, there is an easy proof. $\endgroup$
    – copper.hat
    Jun 2 '20 at 21:49
  • $\begingroup$ Yes, it is $C^2$. $\endgroup$
    – slyyah
    Jun 2 '20 at 21:59
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Here is proof if $f$ is assumed $C^2$. This is not necessary but simplifies the proof significantly.

It is sufficient to show the $f''(x) \ge 0$.

Pick some $x$, then there is a neighbourhood $U$ of $x$ such that $f(x+h) -f(x) \ge f'(x) h$ for $x+h \in U$.

Since $f$ is $C^2$, Taylor gives (for $h$ sufficiently small) that $f(x+h) = f(x) + f'(x)h + {1 \over 2} h^T f''(\xi_h)h$, where $\xi_h \in [x,x+h]$. This gives $h^T f''(\xi_h)h \ge 0$ for $h$ such that $x+h \in U$. If $h \neq 0$ then ${h^T \over \|h\|} f''(\xi_h) {h \over \|h\|} \ge 0$, of course.

Pick some unit vector $v$, and let $h = t v$ for small $t$, then we have $ v^T f''(\xi_{tv}) v \ge 0$, and letting $t \to 0$ and using continuity of $f''$ we get $ v^T f''(x) v \ge 0$.

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  • $\begingroup$ I am curious. Is the statement true, if $f$ is only $C^1$? $\endgroup$
    – user251257
    Jun 8 '20 at 22:23
  • $\begingroup$ @user251257: I believe so. $\endgroup$
    – copper.hat
    Jun 8 '20 at 23:42
  • $\begingroup$ any idea how to proof it? I run out of ideas. $\endgroup$
    – user251257
    Jun 8 '20 at 23:47
  • $\begingroup$ @user251257: Yes, did you ask the question somewhere? $\endgroup$
    – copper.hat
    Jun 9 '20 at 0:36
  • $\begingroup$ now, I did: math.stackexchange.com/q/3711867/251257 $\endgroup$
    – user251257
    Jun 9 '20 at 0:51

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