# If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$

Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by a comment/answer to a different question)

• Proof by picture Jul 15 '13 at 19:45
• This is Barbalat's Lemma (as pointed out by @MrYouMath in a duplicate thread). It only needs $f'$ to be uniformly continuous.
– Dap
Feb 9 '18 at 13:17
• Does the statement still work if we change $\langle \langle$$f'' bounded on \mathbb{R}$$\rangle \rangle$ into $\langle \langle$ $f''$ bounded in a neighborhood of $+\infty$ : $f''(x) \underset{x\to +\infty}{=} O(1)$ $\rangle \rangle$ ? Aug 28 '21 at 0:05

Let $|f''|\le 2M$ on $\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\in\mathbb R$ and every $\delta>0$, there exists $y\in[x,x+\delta]$, such that $$f(x+\delta)=f(x)+f'(x)\delta+\frac{1}{2}f''(y)\delta^2.$$ It follows that $$|f(x+\delta)-f(x)-f'(x)\delta|\le M\delta^2.\tag{1}$$ Since $\lim_{x\to\infty}f(x)=0$, fixing $\delta>0$ and letting $x\to\infty$ in $(1)$, we have $$\limsup_{x\to\infty}|f'(x)|\le M\delta.$$ Since $\delta>0$ is arbitrary, the conclusion follows.

• How is Eq. (1) valid? Can you really take the absolute value for the inequality? Nov 16 '15 at 16:48
• @JoãoVictorBateliRomão he's not taking absolute on the inequality, he is taking it on the equality, and then applying the inequality. Jul 20 '16 at 11:26
• Question: could we get the same results for $f:[0,\infty)\to\mathbb{R}$ and $\lim_{x\to\infty}f(x)=a$ where $a\neq 0$? Jul 20 '16 at 13:27

Let me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to $$\|f'\|^2_{L_{\infty}{\mathbb{(R)}}}\le 4\|f\|_{L_{\infty}{\mathbb{(R)}}}\|f''\|_{L_{\infty}{\mathbb{(R)}}}$$

Let $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\le M|h|$ for all $x,h$. Let $\epsilon>0$ be given. We have to show that $|f'(x)|<\epsilon$ for all sufficiently big $x$. As $f(x)\to0$, there is an $x_0$ such that $|f(x)|<\frac{\epsilon^2}{4 M}$ for all $x>x_0$. Consider $x>x_0$ and assume $f'(x)> 0$. Then $f'(x+h)\ge f'(x)-Mh$ for $h\ge 0$ and hence \begin{align}f\left(x+\frac {f'(x)}{M}\right)-f(x)&=\int_0^{\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\ge \int_0^{\frac {f'(x)}{M}}(f'(x)-M|h|)\,\mathrm dh\\&=\frac{(f'(x))^2}{2M}.\end{align} If on the other hand $f'(x)<0$, we similarly have $f'(x+h)\le f'(x)+Mh$ for $h\ge 0$ and hence \begin{align}f\left(x-\frac {f'(x)}{M}\right)-f(x)&=\int_0^{-\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\le \int_0^{-\frac {f'(x)}{M}}(f'(x)+M|h|)\,\mathrm dh\\&=-\frac{(f'(x))^2}{2M}.\end{align} In both cases we find $$\left|f\left(x+\frac {|f'(x)|}{M}\right)-f(x)\right|\ge \frac{(f'(x))^2}{2M}$$ and as $x+\frac {|f'(x)|}{M}>x_0$, we conclude $\frac{(f'(x))^2}{2M}< 2\cdot \frac{\epsilon^2}{4M}$ and hence $|f'(x)|<\epsilon$ as was to be shown.$_\square$

This is inspired by the proof by picture mentioned in the comment.

Geometrically, if $$f'(x)$$ is large, then because $$f''$$ is bounded, $$f'(x)$$ won't vary too much in a short period, therefore $$f(x)$$ cannot be Cauchy.

Given $$\epsilon>0$$, for sufficiently large $$a$$, due to the convergence of $$f(x)$$ we have $$|f(a+\epsilon)-f(a)| = |\int_a^{a+\epsilon} f'(x)dx|<\epsilon^2$$

By the mean value theorem, $$f'(x) = f'(a) + f''(b)(x-a)$$ for some $$b$$, therefore $$|\int_a^{a+\epsilon} f'(x)dx-\int_a^{a+\epsilon}f'(a)dx|\le M|\int_a^{a+\epsilon}(x-a)dx|$$ Then

$$|\int_a^{a+\epsilon}f'(a)dx|\le |\int_a^{a+\epsilon}f'(x)dx| + M|\int_a^{a+\epsilon}(x-a)dx|\le \epsilon^2 + \frac{M\epsilon^2}{2}$$

Finally $$|f'(a)|\le (1+\frac{M}{2})\epsilon$$

Note that we only need $$f(x)$$ converges (not necessarily to $$0$$) and $$|f''(x)|.