Existence of a minimum for a strictly convex function

Question

Let a function $f : \mathbb R^n \to \mathbb R$ that we assume to be $C^1$ and strictly convex. If we assume that $f$ admits a minimum (thus unique), then is it always true that $f$ "goes to infinity" in all directions, i.e. that $$\{x \in \mathbb R^n \ : \ f(x) \leq M\}$$ is compact for all $M>0$?

Answer 1

Here is a detailed six step argument proving the claim. From that, I am sure, you can find an even simpler proof.

1. Let $p\in\mathbb{R}^n=:D$ the minimum point: $f(p)=m$. Since $f$ is in $C^1$ this means that $f'(p)=0$ (null-vector) and because of strict convexity we have that for any $x\neq p$ that on the straight line $\gamma:I\rightarrow D, s\mapsto (1-s)p+sx$ connecting $p$ and $x$ we have for any $s$ that:

$$f(p) < f((1-s)p+sx) < (1-s)f(p)+sf(x)$$

2. So now consider balls $B^n_r(p)$ centered around $p$ of radius $r$. Specifically their boundary $S^{n-1}_r(p)$. The spheres are closed, bounded sets in a metric space and thus compact. $f, f'$ are continuous in $D$ and so: $$f(p) < f|_{S^{n-1}_r(p)}$$ Since $f$ is continuous there is a minimum at of $f$ on $S^{n-1}_r(p)$. For this minimum we have $f(p) < f(t_r), t_r\in S^{n-1}_r(p), r>0$. Let the increase be $\epsilon_r=f(t_r)-f(p)$.

3. Now lets consider a straight lines from $p$ to $x$ and another point on that line $y$ reached at $0<t=\frac 12<1$, the half-way point.

$$\Delta'=f(x)-f(y), \Delta = f(y)-f(p), \Delta_g=f(x)-f(p)$$

4. Since $f(y)<\frac 12 (f(x)+f(p))=P$ and $f(x)-P=P-f(p)=\frac {\Delta_g}2$ we have that $f(x)-f(y)>f(y)-f(p)$.

5. Fix $\delta>0$. Set $r_n:=n\delta$. We now consider any straight line from $x$ to infinity with $e\in S^{n-1}(0)$ a directional unit vector. Set $x_n:=p+n\delta e$ a sequence of points on the straight line. We know that $f(x_{n+1})-f(x_n) > f(x_1)-f(p) = \epsilon_{\delta} > 0$. Thus by telescoping we have $$f(x_{n+1})-f(p) > n\epsilon_{\delta}$$. Since this holds for all directional unit vectors $e$ we have $\epsilon_{n\delta}>n\epsilon_{\delta}$

6. Now consider $f(x)\leq M$. If $f(p)>M$ then your set is empty and the hypothesis is true. So assume $f(p)\leq M$. Then by our preceding argument:

$$\{x\in D: f(x)\leq M\}\subseteq B^n_{\left(\left\lceil\frac{M-f(p)}{\epsilon_\delta}\right\rceil + 1\right)\delta}(p)$$

Answer 2

Certainly. Consider a sphere of radius 1 around the point $x_0$ where the minimum $m$ is achived, and let $M >m$ the minimum of $f$ on this sphere. Let $x=x_0+t u$ with $t>0$ and $u$ of norm one. then $f(x) > Mt +(1-t)m= (M-m)t +m$, by convexity. As $t= \vert \vert x-x_0 \vert \vert$ you have the result.