How to prove that supremum of strictly convex function is infinity? 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?
 A: 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 $t\to +\infty$.
A: 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.
