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On the interval $(0, \infty)$,the function $f \geq 0$,$f' \leq 0$, and $f'' \geq 0$.Prove that $\lim\limits_{x \to \infty} xf'(x) = 0$.

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3 Answers 3

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If the limit is not zero, there is an $\epsilon>0$ such that $xf'(x) \le -\epsilon$ at arbitrarily large $x$. So we can construct a sequence $x_n$ such that $x_n\ge2^n$, $x_{n+1}\ge 2x_n$ and $x_nf'(x_n) \le -\epsilon$ for all $n$. But then the lower halves of the rectangles that the points $(x_n,-\epsilon/x_n)$ form with the origin all lie above the graph of $f'$ (i.e. between it and the $x$ axis), are all disjoint, and all have area $\epsilon/2$. It follows that the indefinite integral of $f'$, i.e. $f$, diverges to $-\infty$, in contradiction with $f\ge0$.

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It seems like you did not use $f''\geq 0$. Is this correct? –  Fabian Feb 27 '11 at 11:48
    
@Fabian:I did, in saying that the rectangles all lie above the graph of $f'$ -- this is only true because $f'$ can't decrease. If it could, then it could form spikes that extend beyond $-\epsilon/x$ but don't add up to make $f$ diverge to $-\infty$. –  joriki Feb 27 '11 at 12:00
    
Good answer!I get a new proof,maybe f"≥0 is really useless.We get f'(x)<-ε/x,when x>X,(thanks to Shai Covo his answer is also well and show me this idea),fix some y>X,we can do integral from y to 2y,2y to 4y,agian and again,we have f(y)>εlog2+f(2y)>2εlog2+f(4y)>…>nεlog2+f(2^ny)>nεlog2,when n is very large,that is impossible! –  Strongart Mar 1 '11 at 10:28

On the one hand, $f$ is monotone decreasing (since $f' \leq 0$) and $f \geq 0$; hence, for some $l \geq 0$, $\lim _{x \to \infty } f(x) = l$. On the other hand, $f'$ is monotone increasing (since $f'' \geq 0$) and $f' \leq 0$; hence $$ f(x) - f(x/2) = \int_{x/2}^x {f'(u)du} \le \int_{x/2}^x {f'(x)du} = \frac{{x f'(x)}}{2} \le 0. $$ Letting $x \to \infty$, the left-hand side converges to $0$; hence $x f'(x) \to 0$ too.

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For $x>1$, $f(x)$ is bounded below by $0$ and above by $f(1)$. Therefore $\lim_{x\to\infty} {f(x)\over \log x}=0$. By l'Hopital's rule, $\lim_{x\to\infty} xf'(x) =0$ as well.

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This is no way to apply L'Hôpital's rule. From the existence of the limit of $f'/g'$ you may sometimes deduce the existence od the limit of $f/g$, but not the other way around. –  Julián Aguirre Feb 27 '11 at 17:00
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As a counterexample, consider $\lim_{x\to\infty}\frac{\sin x}{x}=0$, whereas $\lim_{x\to\infty}\frac{\cos x}{1}$ does not exist. –  joriki Feb 27 '11 at 19:14

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