Unique global Minimum in a continuous, strictly convex function

Question

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.

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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.

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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)

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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.

Answer 1

1) If $f:[a,b] \rightarrow \mathbb{R}$ is continuous, then $f$ has a global minimum, since $[a,b]$ is compact. Convexity is not needed.

2) Global minimum at $x_0$ means $f(x_0) \le f(x)$ for all $x \in [a,b].$ And not $f(x_0) < f(x).$

A solution to your problem:

Suppose that there are $x_0,x_1 \in [a,b]$ such that $x_0 <x_1 ,$ $f(x_0)=f(x_1)$ and

$$f(x) \ge f(x_0)=f(x_1)$$

for all $x \in [a,b].$ Then there is $t \in [x_0,x_1]$ such that $f(t) \ge f(x_0)=f(x_1).$ ($f$ continuous and $[x_0,x_1]$ compact.)

$f(t)=f(x_0)=f(x_1)$ is not possible, since $f$ is strictly convex. Hence

$$f(t)>f(x_0)=f(x_1),$$

and thus $x_0<t<x_1,$ Hence there is $s \in (0,1)$ with $t=sx_0+(1-s)x_1.$ It follows from the strict convexity that

$$f(t) < sf(x_0)+(1-s)f(x_1)=sf(x_0)+(1-s)f(x_0)=f(x_0),$$

a contradiction.

Answer 2

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.