1
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot. I liked this short proof. But isn't $f(x_1)=f(x_0)$ since both are Minimum? It doesn't change your proof, just asking. $\endgroup$ Mar 12, 2020 at 12:24
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ You're awesome! Thanks for making me realize that I got the definition of a global minimum wrong. But I've a question. Don't you mean $f(x) \geq f(x_0)=f(x_1)$ ? Because otherwise they would be Maxima. $\endgroup$ Mar 12, 2020 at 11:32
  • $\begingroup$ Ooops, you are right. An edit will follow. $\endgroup$
    – Fred
    Mar 12, 2020 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.