I would like to ask for help on this problem...
Show that:
f is a continuous, strictly convex function with $f:[a,b] \rightarrow \mathbb{R}$ $\Longrightarrow$ f has a unique global Minumum.
I've tried a proof by contradiction. So I've to show that it's a contradiction, if
1) f has no global Minimum
2) f has more than one global Minimum.
Starting with 1)
If f has no global Minumum $\Rightarrow$ f has no Minumum at all, because f is bounded in [a,b] $\Rightarrow$ Contradiction to the "Extreme Value Theorem" which states that a continuous function on a closed intervall must have a maximum & minumum.
Going to 2)
If f has more than 2 global Minima, $\Rightarrow$ Contradiction to the definition of a global Minumum ($\forall x \in [a,b]: f(x_0) < f(x)$ with $x_0$ global Minimum)
The problem is: I'm not sure if I've done it right because it seems like I don't need the convex property at all. Can someone proof-read this? Thanks.