Show that if $\lim\limits_{x \to \infty} f(x)$ exists and $f''$ is bounded, then $\lim\limits_{x \to \infty} f'(x)=0$. [duplicate]

I'm trying to answer the following exercise: Suppose that $f$ is twice differentiable on $[0,+\infty)$.

Show that if $\lim\limits_{x \to \infty} f(x)$ exists and $f''$ is bounded, then $\lim\limits_{x \to \infty} f'(x)=0$. If $\lim\limits_{x \to +\infty} f'(x)$ exists then I know how to prove that this limit is equal to 0 (by using the mean value theorem).

But how can I show that this limit exists?

Counterexample: Let us consider $f(x) = x \sin\frac{1}{x}$. Then $$f'(x) = \sin\frac{1}{x} - \frac{1}{x} \cos\frac{1}{x}$$ and $$f''(x) = -\frac{1}{x^2}\cos\frac{1}{x} + \frac{1}{x^2}\cos\frac{1}{x} - \frac{1}{x^3} \sin\frac{1}{x}.$$ This example shows that the hypothesis that there exists $\lim\limits_{x \to \infty} f(x)$ is essential.

marked as duplicate by Guy Fsone, Dap, Jack D'Aurizio limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 9 '18 at 12:51

• This claim is known as Barbalat's Lemma – MrYouMath Feb 9 '18 at 11:43
• I didn't know that. Thanks for this comment! – user85353 Feb 9 '18 at 11:47
• See this. – David Mitra Feb 9 '18 at 12:12
• The question here is different. I want to prove that the limit of $f'(x)$ exists using the boundedness of $f''$. In this link the existence of this limit is hypothesis. – user85353 Feb 9 '18 at 12:12
• Thanks, David Mitra – user85353 Feb 9 '18 at 12:16

We are given that $f''$ is bounded, so let $$|f''(x)|\le M\tag1$$ Suppose that $$\limsup\limits_{x\to\infty}|f'(x)|=A\tag2$$ Since $\lim\limits_{x\to\infty}f(x)$ exists, there is an $L$ so that if $x,y\ge L$, $$|f(x)-f(y)|\le\frac{A^2}{8M}\tag3$$

By $(2)$, there is an $x_0\ge L+\frac{A}{2M}$ so that $|f'(x_0)|\gt\frac A2$. Then, by $(1)$, $|f'(x)|\ge\frac A2-M|x-x_0|$. Since $f'$ has the same sign over $\left[x_0-\frac{A}{2M},x_0+\frac{A}{2M}\right]$, we have \begin{align} \left|\,f\left(x_0+\frac{A}{2M}\right)-f\left(x_0-\frac{A}{2M}\right)\,\right| &=\left|\,\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}f'(x)\,\mathrm{d}x\,\right|\\ &=\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}\left|f'(x)\right|\,\mathrm{d}x\\ &\ge\int_{x_0-\frac{A}{2M}}^{x_0+\frac{A}{2M}}\left(\frac A2-M|x-x_0|\right)\,\mathrm{d}x\\ &=\frac{A^2}{4M}\tag4 \end{align} However, the only way that $(3)$ and $(4)$ do not contradict is if $A=0$. That is, $$\limsup\limits_{x\to\infty}|f'(x)|=0\tag5$$ which implies $$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to\infty}f'(x)=0}\tag6$$

Finiteness of $A$

Note that since $\lim\limits_{x\to\infty}f(x)$ exists, there is an $M_2$ so that for $x\ge M_2$, $|f(x)-f(x+1)|\le1$. The Mean Value Theorem says that for some $\xi\in(x,x+1)$, $|f'(\xi)|\le1$. The Mean Value Theorem and $(1)$ say that for any $\zeta\in(x,x+1)$, $|f'(\zeta)-f'(\xi)|\le M$. Therefore, the $\limsup$ mentioned in $(2)$ is finite; in fact, it is bounded by $M+1$.

• How do you prove the existence of $A$? – Guy Fsone Feb 9 '18 at 16:20
• limsup always exists; it may be infinite, but it exists. I see that in one of my edits, I took out the part that shows it is bounded by $M+1$. I will put that back for clarity. – robjohn Feb 9 '18 at 16:58
• Let assume that $A=\infty$ does that not cause any damage to your computation? see that line before $(4)$ – Guy Fsone Feb 9 '18 at 17:07
• @GuyFsone: why are you worrying about $A=\infty$? I just said I was going to show that $A$ is finite, and I have done so. – robjohn Feb 9 '18 at 17:23
• that is ok now thanks +1) – Guy Fsone Feb 9 '18 at 17:24

Since $f$ is twice differentiable, by Taylor's expansion, for every $\varepsilon>0$, there exists $c_{x,\varepsilon}\in(x,x+\varepsilon)$ such that $$f(x+\varepsilon)=f(x)+f'(x)\varepsilon+\frac{1}{2}f''(c_{x,\varepsilon})\varepsilon^2.$$ But $f''$ is bounded (say, by $|f''|<2M$), so it follows that $$|f(x+\varepsilon)-f(x)-f'(x)\varepsilon|\le M\varepsilon^2.$$ Therefore, by triangular inequality, $$|f'(x)\varepsilon|\le|f(x+\varepsilon)-f(x)|+ M\varepsilon^2.$$ But, $\lim_{x\to\infty}f(x)$ exists, so fixing $\varepsilon>0$ and letting $x\to\infty$, we we get $$\limsup_{x\to\infty}|f'(x)|\le M\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, the conclusion follows.

• This proves that the compose $f' \circ g$, where $g(x)=c_x$, has limit egual to $0$. But does it prove that the limit of $f'$ is $0$? The first equality of the last line is my doubt. – user85353 Feb 9 '18 at 12:03
• @Lucas I have proven the existence now see the edit – Guy Fsone Feb 9 '18 at 12:22
• @user85353 see the edit I have prove the existence and it still works perfectly – Guy Fsone Feb 9 '18 at 12:23