So I am working on the proof that a convex continuous function $f$ on a convex and compact set $X$ attains its maximum at an extreme point of $X\subset\Omega$ where $\Omega$ is a locally convex space and there is a small part that is tripping me up. So far I have:

Suppose, for contradiction, that $f$ attains its max at a point that is not extreme call it $x_0$. So there exists $x_1,x_2\in X$ distinct and $t\in(0,1)$ s.t. $x_0=tx_1+(1-t)x_2$. This means by the convexity of $f$ that $$f(x_0)\leq tf(x_1)+(1-t)f(x_2)\leq\max\{f(x_1),f(x_2)\},$$ which yields a contradiction when the strict inequality holds. My problem is getting the result when equality holds. I still haven't used continuity of $f$ or compactness of $A$ and so I am thinking those must come into play here, but I can't see how.


There's no issue with the function attaining its maximum at a non-extreme point, so long as it also attains its maximum at an extreme point. For example, constant functions will achieve its maximum at all points!

What the argument you've provided shows is that the set of points on which $f$ achieves its maximum is a face of $X$. That is, if $x_0$ maximises $f$ over $X$, and $x_0 \in [x_1, x_2]$, then $f$ is maximised at $x_1$ and $x_2$ as well.

Moreover, since $f$ is continuous, this face is closed, and hence compact. By the Krein-Milman Theorem, this face must contain an extreme point (of the face). It's not hard to see that such an extreme point of the face must be an extreme point of $X$. Thus, $f$ always achieves its maximum on an extreme point.

  • $\begingroup$ Okay, I think that makes sense. To edit my proof could I just choose $x_1,x_2$ to be extreme points? $\endgroup$ – Scott Feb 22 at 2:46
  • $\begingroup$ No, not necessarily. Not every point is on the line segment of two extreme points (e.g. look at non-degenerate triangles in $\Bbb{R}^2$). Moreover, I don't think it's clear a priori that you can assume $x_1$ or $x_2$ is an extreme point. $\endgroup$ – Theo Bendit Feb 22 at 3:07

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