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Suppose that $f \in L^1(0,+\infty)$ is a monotonic function. Prove that $\lim_{x \to +\infty} x f(x)=0$.

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If this a homework question - please add the homework tag and explain what you've tried so far. – nbubis Apr 22 '12 at 9:40
Actually it isn't a homework question. I believe it pertains to some enrollment test for a Ph.D. program, but I'm not sure. – Siminore Apr 22 '12 at 9:43
Of course, the main issue is the existence of the limit. If the limit exists, then it must be zero. I guess that it suffices to prove that $\limsup_{x \to +\infty} x f(x)=0$. – Siminore Apr 22 '12 at 9:44
when x goes to infinity,maybe we can start from -infinity to plus infinity,and because f(x) is always positive,it would canceal out each other right yes? – dato Apr 22 '12 at 9:46
Well, the function is not defined on the whole real line. I think the solution should be more... tricky. – Siminore Apr 22 '12 at 9:50

1 Answer

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We can assume WLOG that $f$ is decreasing. Then $f$ is non-negative, otherwise, if $f(x_0)<0$ then $f(x)\leq f(x_0)<0$ for $x\geq x_0$ and $f$ wouldn't be integrable. We can write $$|xf(2x)|=\left|\int_x^{2x}f(2x)dt\right|=\int_x^{2x}f(2x)dt\leq \int_x^{2x}f(t)dt=\int_0^{2x}f(t)dt-\int_0^xf(t)dt.$$ Since $f$ is integrable, the RHS converges to $0$, so $\lim_{x\to 0}2xf(2x)=0$ and we are done.

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This is a really nice proof: elementary and yet tricky! I think that a more standard proof could be found, based on something like Riemann sums and "limsup" estimates. But yours is surely better than this. – Siminore Apr 22 '12 at 13:49

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