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Let $f$ be a $C^\infty(\mathbb R,\mathbb R)$ function.

I'm reading a proof where the author bluntly states the following:

Since $\lim_{x\to \infty}f(x)=L$ and $f''$ is bounded, $\lim_{x\to \infty}f'(x)=0$

Since no proof is given, I'm assuming this is something basic, but I haven't found a proof.

Consider two reals $x$ and $x_0$. From $\displaystyle f(x)=f(x_0)+f'(x_0)(x-x_0)+\int_{x_0}^x (x-t)f''(t) dt$ I derive $$|f'(x_0)|\leq \left|\frac{f(x)-f(x_0)}{x-x_0}\right|+ \sup \left|f''\right|\frac{|x-x_0|}2$$

If I let $x\to \infty$, the RHS goes to $\infty$ and the proof is ruined...

If I choose $x$ close to $x_0$, the inequality looks like $$\sup \left|f''\right|\frac{|x-x_0|}2\geq 0$$

Not good...

Gabriel Romon
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2 Answers2

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If $f'$ does not tend to $0$ then $|f'(x_n)|\ge\epsilon>0$ for some $x_n\to\infty$. Wlog $f'(x_n)\ge\epsilon$. Since $f''$ is bounded this implies there exists $\delta>0$ such that $f'>\epsilon/2$ on $(x_n-\delta,x_n+\delta)$, which then shows $\lim f(x)$ cannot exist.

David C. Ullrich
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Here's a quantitative bound leading to the result (inspired by Exercise 15 in Chapter 5 of Rudin's PMA). Let $M$ be a bound for $f''$. Because $\lim_{x\to\infty} f(x)=L$ exists, given $\epsilon>0$ there exists $N$ so large that if $x,y\ge N$ then $|f(x)-f(y)|<\epsilon$. Fix $x\ge N$,and let $h>0$. By Taylor's theorem, there exists $\xi\in(x,x+h)$ such that $$ f(x+h)=f(x)+hf'(x)+(h^2/2)f''(\xi). $$ Isolating $f'(x)$ and making the obvious estimates we obtain $$ |f'(x)|\le {\epsilon\over h}+{Mh\over 2}. $$ Optimizing the choice of $h$ ($h=\sqrt{2\epsilon/M}$) yields $$ |f'(x)|\le\sqrt{2\epsilon M}, $$ provided $x\ge N$.

John Dawkins
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