Suppose $f$ is a twice differentiable function such that $f''(x) \ge 0$ for all $x \in \mathbb R$. Let $0 \le f(x) \le 1$ for $x \ge 0$. Then show that $\lim_{x \rightarrow \infty} f(x)$ exists.

I have tried it but I fail. Please give me some hint to proceed in the right way.

Thank you in advance.


$f''>0$ therefore $f'$ is increasing.

  • First case: $\forall x, f'(x)\le0$. Then $f$ is always decreasing, and since it's bounded it converges.

  • Second case: Likewise, $\forall x, f'(x)\ge0$. Then $f$ is always increasing, and since it's bounded it converges.

  • Third case: $f'$ is negative until some $x_0$ where it becomes positive (remember, $f'$ is increasing). Then we're back to case 2 and $f$ converges.


As pointed out by @MatthewLeingang, cases 2 and 3 are not actually necessary. Indeed, a function that verifies $f''(x)>0,f'(x)>0$ for all $x\ge x_0$ cannot converge.

This can be proved easily: suppose that $\lim_{x\to\infty}f(x)=l<\infty$. By definition for any $\epsilon>0$ there exists $x_1\ge x_0$ such that $|l-f(x_1)|=l-f(x_1)<\epsilon$. ($l>f(x_1)$ since $f$ is increasing)

However, since $f'$ is also increasing, we have $$f(x_1+\frac{2\epsilon}{f'(x_1)})\ge f(x_1)+2\epsilon>l$$ which is absurd. Hence $f$ does not converge.

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  • 1
    $\begingroup$ Do the second and third case ever happen? I was thinking a convex increasing function would have to have a limit of $\infty$. $\endgroup$ – Matthew Leingang Jun 13 '17 at 16:11
  • $\begingroup$ @MatthewLeingang Indeed when at some point $f'(x)>0$ then one can show that $f$ can not possibly converge and therefore diverges to $\infty$. That, however, is not needed for our proof here. $\endgroup$ – Hippalectryon Jun 13 '17 at 16:18
  • $\begingroup$ True, it doesn't change the validity of the proof. But a proof by cases where two of the three are vacuous is somewhat...inelegant. $\endgroup$ – Matthew Leingang Jun 13 '17 at 16:22
  • $\begingroup$ @MatthewLeingang Alright, I added it in the answer. Thanks. $\endgroup$ – Hippalectryon Jun 13 '17 at 16:30
  • $\begingroup$ @Hippalectryon I think for the first case you consider we have $\lim_{x \rightarrow \infty} f(x) = \underset {x \in (-\infty,a)} {\inf} f(x)$ for any $a \in \mathbb R$, for the second case we similarly have $\lim_{x \rightarrow \infty} f(x) = \underset {x \in (a,\infty)} {\sup} f(x)$ for any $a \in \mathbb R$ and for the third and final case we have $\lim_{x \rightarrow \infty} f(x)=\underset {x \in (x_{0},\infty} {\sup} f(x)$ for some $x_{0} \in \mathbb R$ what you have mentioned in your answer. $\endgroup$ – Arnab Chatterjee. Jun 14 '17 at 7:15

A brief answer.

If $f:\mathbb R\to\mathbb R$ is convex, then exactly one of the following three holds:

a. $f$ is increasing,

b. $f$ is decreasing,

c. There exists an $x_0\in\mathbb R$, such that $f$ is decreasing in $(-\infty,x_0]$ and increasing in $[x_0,\infty)$.

In all three cases, $f$ is eventually monotonic (i.e., it is monotonic in some interval $[a,\infty)$, for some $a\ge 0$). Since $f$ is also bounded, in the same interval, then the limit $\lim_{x\to\infty}f(x)$ exists, and it lies in the interval $[0,1]$, since $0\le f(x)\le 1$, for $x\ge 0$.

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