Suppose that $f \in L^1(0,+\infty)$ is a monotonic function. Prove that $\lim_{x \to +\infty} x f(x)=0$.
Integrable monotonic functions
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calculus
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0Well, the function is not defined on the whole real line. I think the solution should be more... tricky. – 2012-04-22
1 Answers
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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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0This 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. – 2012-04-22