# Limit of a convex function

I would need a check on the following exercise:

Let $$f:\mathbb{R} \rightarrow \mathbb{R}$$ a convex function.

• Prove that $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow - \infty} f(x)$$ exist

• Show that if both the limits are finite, then $$f$$ is constant.

My attempt:

i ) I know that if $$f$$ is convex, then $$f(t x_1 + (1-t)x_2)< t f(x_1) + (1-t) f(x_2)$$

If I fix an arbitrary $$N>0$$, then I have that for $$x>x_2 \colon \quad f(x)>N$$, thanks to the convexity, therefore this proves the limit to $$+ \infty$$ is $$+\infty$$.

The same argument applies to $$\lim_{x \rightarrow -\infty}f(x)$$: it suffices to note that for $$xN$$.

ii)

Graphically it's obvious, but I have some problem in make it formal.

If the limit is finite, say $$L$$, then for every $$\varepsilon >0$$ there exists an $$M(\varepsilon)$$ such that for $$x>M(\varepsilon) \colon \quad |f(x)-L|\leq \varepsilon$$

Assume $$f (x) \ne c$$. By definition of convexity, it has to hold (for $$t \in [0,1]$$) $$t f(M)+(1-t)f(M+1) \leq f(t M + (1-t)(M+1))$$

Now, by definition of limit, $$f(M)$$ and $$f(M+1)$$ are less than $$L-\varepsilon$$. Also, the argument in the rhs of the inequality can be simplified:

$$L-\varepsilon

Therefore $$L-\varepsilon < f(M-t)$$, which is a contradiction because $$M-t and hence it can be greater than $$L-\varepsilon$$.

So $$f$$ has to be equal to $$c$$. Indeed in this case, it is still (trivially) convex, and the limits are of course finite.

• Your argument for (i) seems unsatisfactory to me Aug 18, 2020 at 8:55
• Thanks for the check. How could I improve it? (do you think that (ii) is fine)? @AdamRubinson Aug 18, 2020 at 8:55
• You stated that, "for $x > x_2, f(x) >N$, due to it being convex". But that's not a mathematical proof. That's just an assertion without evidence/reason. Aug 18, 2020 at 9:02
• @AdamRubinson you're right, I was thinking about the graphical representation of a convex function. The fact is that I don't know how to formalize it. Could you give a hint? Aug 18, 2020 at 9:06
• Yeah I don't know the answer. It's not super easy. But I'll give it some thought... Aug 18, 2020 at 9:18

convexity usually means "$$\le$$", not "$$\lt$$" (otherwise it's "strictly convex").

You don't want to show that f always goes to infinity, because it need not.

Start with $$x\rightarrow\infty$$.

Suppose first there are two points $$x\lt y$$ with $$f(x)\lt f(y)$$. Then we can show that f goes to infinity. We can suppose without loss of generality that $$x=0$$ and $$f(x)=0$$ (if not, just slide and shift f until it does. It won't change the behaviour we are interested in.)

Consider some $$z>y$$. Since $$y>x=0$$, then $$z=y/t$$ for some $$0. So by convexity, $$tf(z)=tf(y/t)+(1-t)f(0)\ge f(t(y/t)+(1-t)0)=f(y)$$ So $$f(z)\ge f(y)/t$$. As $$z\rightarrow \infty$$ it's clear that $$t$$ goes to $$0$$, so $$f(y)/t\rightarrow\infty$$ (remember $$f(y)>0$$) and therefore so does $$f(z)$$. Therefore in this case $$f$$ increases to infinity.

Otherwise our supposition was false, so f must be either constant or else non-constant and monotone decreasing. Suppose it's the latter. Again, move the origin so that $$f(0)=0$$. Then $$f(1)<0$$ and it's easy to show by the convexity property that $$f(t)\le t f(1)$$ and so $$f$$ goes off to minus infinity.

You can then repeat the argument by symmetry, for the behaviour as $$x\rightarrow-\infty$$.

Hint : try to prove that a convex function is either decreasing, either increasing, either decreasing then increasing.