19
$\begingroup$

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)

$\endgroup$
  • 2
    $\begingroup$ Proof by picture $\endgroup$ – 40 votes Jul 15 '13 at 19:45
  • $\begingroup$ This is Barbalat's Lemma (as pointed out by @MrYouMath in a duplicate thread). It only needs $f'$ to be uniformly continuous. $\endgroup$ – Dap Feb 9 '18 at 13:17
29
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How is Eq. (1) valid? Can you really take the absolute value for the inequality? $\endgroup$ – João Victor Bateli Romão Nov 16 '15 at 16:48
  • $\begingroup$ @JoãoVictorBateliRomão he's not taking absolute on the inequality, he is taking it on the equality, and then applying the inequality. $\endgroup$ – Michael Jul 20 '16 at 11:26
  • $\begingroup$ Question: could we get the same results for $f:[0,\infty)\to\mathbb{R}$ and $\lim_{x\to\infty}f(x)=a$ where $a\neq 0$? $\endgroup$ – Michael Jul 20 '16 at 13:27
8
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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.$_\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.