Maximum and supremum: proving sup of a concave function is < than infinity

Question

Statement: Fix any $a \leq b$ in $\mathbb{R}$. If $f \in \mathbb{R}^{[a,b]}$ is concave (or convex), then $\sup$ $f([a,b])$ $< \infty$.

Wouldn't that be automatically implied by the Weierstrass theorem?

Weierstrass: If $f \in \mathbb{R}^{[a,b]}$ is continuous, then $\exists x, x' \in [a,b]$ such that $ \forall t \in [a,b]$, $f(x') \leq f(t) \leq f(x)$.

Since $f(x)$ is a maximum, wouldn't it also be the supremum of the set $f([a,b])$?

I know there is a proof of the first statement that isn't as simple as "implied by Weierstrass". So my question is why isn't it implied by Weierstrass?

Thank you

Answer 1

Yes, what you say is clear from Weierstrass, without assuming concavity. Unless you're not given that $f$ is continuous.

It seems possible that you're stating the result incorrectly. In fact if you just know that $f$ is concave on $(a,b)$ then it follows that $\sup_{(a,b)}f<\infty$, and this is not just Weierstrass...

Answer 2

Concave/convex functions are not necessarily continuous. They are continuous in the realtive interior of their domain, but they need not be on its boundary. For instance, the function $g(x)=\begin{cases}0&\text{if }a<x<b\\ 1&\text{if }x=a\vee x=b\end{cases}$ is convex on $[a,b]$ - and thus continuous on $(a,b)$ -, but not continuous on $[a,b]$. Of course, this begs for a specific argument to justify why a convex function on $[a,b]$ must be bounded on $(a,b)$. Namely: why doesn't $\tan x$, which is convex and unbounded on $\left[0,\frac\pi2\right)$, have a convex extension to $\left[0,\frac\pi2\right]$?