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Suppose there is a strictly convex continuous function $f$: $R^n$ $\rightarrow$ $R$.

Is the supremum of $f$ always infinity? How can we prove it?

I am trying to come up with proof. If $x$ and $y$ are two points in $R^n$, strictly convex implies $f(\alpha x_1 + (1-\alpha) x_2) $ < $\alpha f(x_1) + (1-\alpha)f(x_2)$ . Suppose $f$ is bounded.

Case 1: The bound is attained at a point, say $x_0$. Then for some $\alpha$, some $x_1$ and $x_2$ s.t. $ (\alpha x_1 + (1-\alpha) x_2) = x_0$:
$f(x_0)$< $\alpha f(x_1) + (1-\alpha)f(x_2)$
Therefore a contradiction.
Case 2: The bound is not attained. Since the function is strictly convex, we know $f(x)$ approaches this bound as $x$ approaches $ \infty $
I don't know how to proceed after this step. Where can I find a contradiction in this case?

Tushar Pandey
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2 Answers2

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You can prove the following result:

Claim: let $f\colon\mathbb{R}^n\to\mathbb{R}$ be a convex function. If $f$ is bounded from above, then $f$ is constant.

To prove the claim, assume by contradiction that there exist two points $x_0, x_1$ such that $f(x_0) < f(x_1)$, and consider the function $$ \varphi(t) := f((1-t)x_0 + t x_1), \qquad t\in\mathbb{R}. $$ Since $\varphi$ is convex, we have that $$ \varphi(t) \geq (1-t)\varphi(0) + t \varphi(1), \quad \forall t\geq 1, $$ i.e. $$ f((1-t)x_0 + t x_1) \geq f(x_0) + t [f(x_1) - f(x_0], \quad\forall t\geq 1. $$ But the r.h.s. goes to $+\infty$ as $t\to +\infty$.

Rigel
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  • Thank you. However, I think you proved the statement directly, not by contradiction. – Tushar Pandey Jul 12 '20 at 23:38
  • @Tushar Pandey The contradiction is that $f$ is unbounded. There seem to be some other issues however. For example that convexity does not imply that inequality (at least not directly). – L. t. May 25 '21 at 00:35
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I am sure this has been answered before but am unable to find a solution.

The key result is that if $f$ is a non constant convex function on $\mathbb{R}^2$ then there are $x,y$ such that $f(x)<f(y)$. Let $\phi(t) = f(x + t(y-x))$ and suppose $s >1$. Then we can write $1={1 \over s} s + (1 - {1 \over s}) 0$ and so $\phi(1) \le {1 \over s} \phi(s) + (1 - {1 \over s}) \phi(0)$, or $\phi(s) \ge \phi(0) + s(\phi(1) - \phi(0)) = f(x) + s (f(y)-f(x))$ and so $f$ is unbounded.

If $f$ is strictly convex then it must be non constant.

copper.hat
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  • Thank you. Looks like I was approaching the problem in a different manner which doesn't work with $R^n$ but works only with R (since there is a natural order in R) – Tushar Pandey Jul 12 '20 at 23:37