# If $f$ increasing, analytic on $\mathbb{R}$ and $\lim_{x\to +\infty}f(x)=1$, does it follows that $\lim_{x\to +\infty}f'(x)=0$?

Question:

If $$f$$ strictly increasing, analytic on $$\mathbb{R}$$ and $$\lim_{x\to +\infty}f(x)=1$$, does it follows that $$\lim_{x\to +\infty}f'(x)=0$$?

If we drop the assumption that the function is increasing, an easy counterexample is $$a(x)=\frac{\sin(x^2)}{x}+1$$.

If we drop the analyticity requirement (but keep $$C^{\infty}$$) a counterexample can be constructed from

$$h(x)=\begin{cases}0&x\le 0\\\exp\left(\frac{-\exp(-1/{(x-1)^2})}{x^2}\right)&x\in (0,1)\\ 1&x\ge 1\end{cases}$$

by setting

$$b(x):=\text{sign}(x)\sum_{n=0}^{+\infty}\frac{h(2^n(|x|-n))}{2^{n+1}}$$

It is clear that, if $$\lim_{x\to +\infty} f'$$ exists, it must be $$0$$:

In fact, since $$0=\lim_{x\to +\infty}\frac{f(x)-1}{x}=\lim_{x\to +\infty}f'(x)$$.

Otherwise one can prove it by noting that, since $$f'\ge 0$$ and $$1=\int_0^\infty f'(x)dx$$ it is impossible to have $$\lim_{x\to +\infty}f'(x)>0$$.

However, I do not see how to prove the existence of the limit of how to construct a counterexample (as $$b$$ is not analytic)

• Are you sure it is wrong ?
– EDX
May 1, 2020 at 13:40

$$f(x) = \int_0^x \exp(-t^4 \sin^2(t))\, dt$$

is a counterexample, after normalizing by a suitable constant so that $$\lim_{x \to \infty} f(x) = 1$$ instead of $$1.17195 \ldots$$

The idea is that the factor $$t^4$$ suppresses the contribution of the integrand at all points too far from the zeroes of $$\sin(t)$$ as $$t$$ gets large, and $$\sin^2(t)$$ is only within $$O(1/N^4)$$ of $$0$$ for $$t$$ that's $$O(1/N^2)$$ from an integer multiple of $$\pi$$. Since $$1/N^2$$ is a fast enough decaying sequence, the integral converges. It's obviously an analytic function.

Edit: Here is a more formal argument. Consider the integral

$$\int_{N \pi}^{(N+1) \pi} \exp(-t^4 \sin^2(t)) \, dt \leq \int_0^{\pi} \exp(-\pi^4 N^4 \sin^2(t))$$

Split the interval $$[0, \pi]$$ into two pieces, $$[1/N^{5/4}, \pi - 1/N^{5/4}]$$ and its complement. Over this interval,

$$\sin^2(t) \geq \sin^2(1/N^{5/4}) = 1/N^{5/2} + O(1/N^5)$$

so that $$\exp(-\pi^4 N^4 \sin^2(t)) \leq \exp(-\pi^4 N^{3/2} + O(N^{-1}))$$, which is obviously summable by comparison to the geometric series, say. So the contribution to the integral from this part is $$L^1$$. The other part of the integral is over a set of measure $$2/N^{5/4}$$, and since the integrand is bounded from above by $$1$$, this part only contributes a term of order $$O(1/N^{5/4})$$, which is also $$L^1$$. We conclude therefore that $$f$$ is well-defined by monotonicity and has a finite limit at infinity.

• Why is it an analytic function? May 1, 2020 at 13:48
• Just need to be a bit more formal for the conclusion I think, but this is brilliant
– EDX
May 1, 2020 at 13:48
• So in this case, the limit $f'(x)$ does not exist. Suppose the original post had made the assumption that $f'(x)$ existed. May 1, 2020 at 13:48
• @ClementYung because the integrand is analytic.
– EDX
May 1, 2020 at 13:49
• @user2379888 They didn't actually ;)
– EDX
May 1, 2020 at 13:49

I was thinking we could write the limit as $$\lim_{x \to \infty}\frac{e^xf(x)}{e^x}=1$$ Since this is an indeterminate form, we can apply L'Hopital's Rule, $$\therefore \lim_{x \to \infty}\frac{e^xf(x)}{e^x}=\lim_{x \to \infty}\frac{e^xf(x)+e^xf'(x)}{e^x}=1$$ $$\therefore \lim_{x \to \infty}(f(x)+f'(x))=1 \implies \lim_{x \to \infty}f'(x) = 0$$

If I have made some error, please feel free to correct me.

• L'hopital works only if the limit of the derivative exists, which is the hard part to prove here
– user515010
May 1, 2020 at 13:47
• Ahhh, I missed that. I thought when you said that the function is analytic, it will have a derivative and its limit will exist :) May 1, 2020 at 13:51