Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be continuous such that $\lim\limits_{\|x\| \to\infty} f(x)=\infty$, then there exist a global minimum.

Question

Statement*:

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be continuous such that $\lim\limits_{\|x\| \to\infty} f(x)=\infty$, then there exist a global minimum.

Note: $\|.\|$ is the euclidean norm.

Proof:

Because $f(x)$ is diverging to infinity if $\|x\| \to\infty$ there exists $K > 0$ for every $M > 0$, such that if $x\in \mathbb{R}^n$ with $\|x\|>K$ then $f(x) > M$ and if $x\in \mathbb{R}^n$ with $\|x\|\leq K$ then $f(x) \leq M$.

Define $S^{n-1}=\{x\in \mathbb{R}^n |x_1^2+...+x_n^2\leq K^2\}$ then $S^{n-1}$ is compact because it is bounded and closed.

So the restriction of $f$ to the compact set $S^{n-1}$ has a global minimum and maximum because $f|_{S^{n-1}}$ is continuuous. Let $a\in S^{n-1}$ be the point in which $f|_{S^{n-1}}$ has its global minimum $f(a)$ then $\|a\|\leq K$ and $f(a)\leq M$.

For any $x\in \mathbb{R}^n\setminus S^{n-1}$, $\|x\|> K$, and thereby $f(x) > M\geq f(a)$.

So $f(a)$ is a global minimum.

q.e.d.

*The statement is almost the same as in the following question:

Limit goes to infinity, show that the f has a finite minimum.

I want to adapt the proof given in the answer for the domain $\mathbb{R}^n$ instead of $\mathbb{R}$ and show that there is a global minimum.

Did I made any mistakes?

Answer 1

It's not true that you can choose $K$ s.t. if $\|x\| \leq K$ then $f(x) \leq M$ - that's true only if $f(x) = g(\|x\|)$ where $g$ is monotonic, and most functions are not of such form.

The main idea is correct, however. Just take $M = f(0)$, choose $K$ s.t. if $\|x\| > K$ then $f(x) > M$. Then, as you did, choose $a$ s.t. $f|_{S^{n-1}}$ has global minima at $a$, and now we indeed have $f(a) \leq f(0) = M$.