Given convexity, $\lim_{\|x\| \xrightarrow{} \infty} f(x) = \infty$ is equivalent to $\liminf_{\|x\|\xrightarrow{} \infty} \frac{f(x)}{\|x\|} > 0$?

Question

Source: This is the part of Theorem 4.11 left as an exercise from this book https://link.springer.com/book/10.1007/978-3-031-02406-1

Let $f: \mathbb{R}^{n} \xrightarrow{} \mathbb{R} \cup$ {$\infty$} be convex and let $\inf \{f(x): x \in \mathbb{R}^{n} \} < \infty$. Then the following are equivalent:

(i) $\lim_{\|x\| \xrightarrow{} \infty} f(x) = \infty$.

(ii)$\liminf_{\|x\|\xrightarrow{} \infty} \frac{f(x)}{\|x\|} > 0$.

I cannot get (ii) from (i), because I don't know how to find a bounded sequence $f(x), \|x\| \xrightarrow{} \infty$. And I don't know how to use convexity. My attempt is as follows:

Since $\lim_{\|x\| \xrightarrow{} \infty} f(x) = \infty$, fix $M > 0$, there is $\delta_{0} > 0$ s.t. $\|x\| \geq \delta$ implies $f(x) > M$, which means $\inf_{\|x\| \geq \delta_{0}} \frac{f(x)}{\|x\|} \geq 0$.

If (ii) doesn't hold, then this means $\lim_{\delta \geq \delta_{0}, \delta \uparrow \infty} \inf_{\|x\| \geq \delta} \frac{f(x)}{\|x\|} = 0$. But as $\delta$ increases, $\inf_{\|x\| \geq \delta} \frac{f(x)}{\|x\|}$ increases. So for any $\delta \geq \delta_{0}$, $\inf_{\|x\| \geq 0} \frac{f(x)}{\|x\|} = 0$.

Now pick a sequence $(\delta_{n}) \uparrow \infty$ starting from the above $\delta_{0}$. Then there is an $x_{i}$ s.t. $\|x_{i}\| \geq \delta_{i}, \frac{f(x_{i})}{\|x_{i}\|} \leq \epsilon_{i}$. Now I don't know how to pick $\epsilon_{i}$ and even I can, it seems the direction of the inequality (norm of $x_{i}$) is not what I want.

Answer 1

Here is a proof of (ii) implies (i).

Take $x_0$ with $f(x_0)<+\infty$. Due to (i), for $M:=f(x_0)+1$ there is $R$ such that $f(x)>M$ for all $x$ with $\|x\|\ge R$.

Fix $x$ with $\|x\|= R$. Then also $f(x) \ge f(x_0)+ 1$. Take $s>1$. Define $$ x_s = s x + (1-s)x_0, $$ so that $x = s^{-1} x_s + (1-s^{-1})x_0$. By convexity $$ f(x) \le s^{-1} f(x_s) + (1-s^{-1})f(x_0), $$ or equivalently $$ f(x_s) \ge s f(x) +(1-s) f(x_0) = s ( f(x)-f(x_0)) + f(x_0) \ge s + f(x_0). $$ In addition $$ \|x_s\| = \| s(x-x_0) + x_0 \|\le 2R s + \|x_0\|. $$ Hence $$ \frac{ f(x_s)-f(x_0) }{\|x_s\|} \ge \frac{ s }{2R s + \|x_0\|} \ge \frac{ 1 }{2R + \|x_0\|}. $$ Every $x$ with $\|x\|>R$ can be written as such an $x_s$, so that $$ \frac{ f(x)-f(x_0) }{\|x\|}\ge \frac{ 1 }{2R + \|x_0\|}\quad \forall x: \|x\|>R. $$

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Here is a first attempt to prove it using a contradiction. I needed to assume that $f$ is lower semicontinuous.

Assume (ii) does not hold. Then there is a sequence $x_n$ such that $\|x_n\|\to\infty$ and $\lim \frac{ f(x_n)}{\|x_n\|}=0$. By extracting a subsequence if necessary, we can assume that $u_n := \frac1{\|x_n\|}x_n$ converges to some $u$. Note that $\|u\|=1$.

Now let $x_0$ such that $f(x_0)<\infty$. We get by lower semicontinuity $$ f( x_0 + u) \le \liminf f(x_0 + \frac1{\|x_n\|}(x_n -x_0)). $$ For $n$ such that $\|x_n\|\ge 1$, we get $$ f(x_0 + \frac1{\|x_n\|}(x_n -x_0)) \le (1-\frac1{\|x_n\|})f(x_0) + \frac1{\|x_n\|} f(x_n) \to f(x_0), $$ so that $$ f( x_0 + u) \le f(x_0)< \infty.$$ Now we can repeat the argument for $x_0+u$ instead of $x_0$. By induction it follows that $$ f(x_0 + k u) \le f(x_0)\quad \forall k\in \mathbb N, $$ which contradicts (i).

Answer 2

This is my attempt to show the easier direction.

$\liminf_{\lVert x \rVert \xrightarrow{} \infty} \frac{f(x)}{\lVert x \rVert} = c > 0$. Fix $M > 0$ arbitrarily large, and $\epsilon > 0$ arbitrarily small. Since $\inf_{\lVert x \rVert \geq \delta} \frac{f(x)}{\lVert x \rVert}$ increases as $\delta$ increases, there is $k>0$ s.t. $c \geq \inf_{\lVert x \rVert \geq k} \frac{f(x)}{\lVert x \rVert} \geq c(1-\epsilon)$.

Now take $M' = \max\{k, M/c\}$. So $c \geq \inf_{\lVert x \rVert \geq M'} \frac{f(x)}{\lVert x \rVert} \geq c(1-\epsilon).$ So for all $\lVert x \rVert \geq M', $ we have that $f(x) \geq c(1-\epsilon) \lVert x \rVert \geq c(1-\epsilon)M' \geq c(1-\epsilon)M/c = M(1-\epsilon)$. This means $f$ goes to infinity as the norm of $x$ does so.

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I asked someone and the 1D case is solved for the other direction. Suppose given (i), (ii) doesn't hold.

$f$ goes to infinity as $\lVert x \rVert$ goes to infinity so we consider the $x$'s in the space whose $f$'s are positive.

Since $\liminf \frac{f(x)}{\lVert x \rVert} = 0$, there is a sequence $(x_{n})$ such that $\frac{f(x_{n})}{\lVert x_{n} \rVert} \downarrow 0$ and $\lVert x_{n} \rVert \xrightarrow{} \infty$. Up to subtracting a subsequence of the same sign we assume all $(x_{n})$ are positive. We show $(f(x_{n}))$ must be bounded, so (i) doesn't hold.

Suppose $(f(x_{n}))$ is unbounded, then up to subtracting a subsequence, we can assume $(f(x_{n})) \uparrow \infty$. However, now we have that the slope $c_{n} := \frac{f(x_{n})}{\lVert x_{n} \rVert}$ is decreasing but $f(x_{n})$ increases to infinity. Connecting the dots $(x_{n}, f(x_{n}))$ we find the graph is concave, a contradiction.