Unique global Minimum in a continuous, strictly convex function

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.

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 $$x_0). Let $$\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.

• 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. Mar 12, 2020 at 12:24

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 $$f(x_0) < f(x).$$

Suppose that there are $$x_0,x_1 \in [a,b]$$ such that $$x_0 $$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 $$x_0 Hence there is $$s \in (0,1)$$ with $$t=sx_0+(1-s)x_1.$$ It follows from the strict convexity that

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

• 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. Mar 12, 2020 at 11:32