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.
Best Answer
Proof (1) is not precise but the idea is correct. For (2), I don't see any proof.
Let $f$ strictly convex, and suppose that there are two global minimums at $x_0$ and $x_1$ (where $x_0<x_1$). Let $\lambda \in (0,1)$. Then $$f(x_0)\leq f\big(\lambda x_0+(1-\lambda )x_1\big)< \lambda f(x_0)+(1-\lambda )f(x_1)$$
$$\underset{f(x_1)\leq f(x_0)}{\leq} \lambda f(x_0)+(1-\lambda )f(x_0)=f(x_0),$$ which is a contradiction.