Suppose that $f\in L^1(0,+\infty)$ is a monotone decreasing, positive function. Prove that $$\lim_{x \to +\infty}x(\log x)\cdot f(x)=0.$$
Limit of $x\log x\cdot f(x)$ when $f$ is integrable function
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1xln(x)f(x) $\qquad$ or $\qquad$ xln(xf(x))?? – 2012-12-02
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0$f(x) x \ln x $ – 2012-12-02
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0The problem is to see whether there is a sequence $\{t_k\}$ of real numbers, increasing to $+\infty$ and such that the series $\sum_k\frac{t_k-t_{k-1}}{t_k\log t_k}$ is convergent. – 2012-12-02
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1@DavideGiraudo Good analysis. There is. – 2012-12-02
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0Yes, for example $t_k=2^{k^2}$. – 2012-12-03
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1OP: To modify drastically the question after some answers are posted is contrary to the policy of the site (and to politeness, I should add). Please do not do this. If you have a question different from the one posted, then post another question. (I reverted your question to its previous version.) – 2012-12-05
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0I am very sorry. I tried to post another question but the system told me that it is the same as the above one. I would like to post now the question with the additional condition, f to be convex. Could anybody help me to post the new question? – 2012-12-07
3 Answers
The assertion is not true. Consider $$ f_0=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\mathbf 1_{[0,\mathrm e^{n^2}]}, $$ then $f_0$ is nonincreasing, $g_0(x)=f_0(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g_0(\mathrm e^{n^2})\geqslant1$ for every $n\geqslant1$, and the integral of $f_0$ is $\sum\limits_{n\geqslant1}n^{-2}$, which is finite, hence $f_0$ is integrable.
One can modify this example slightly to get a decreasing function $f$, for example $$ f(x)=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\exp(-x\mathrm e^{-n^2})\mathbf 1_{[0,\mathrm e^{n^2}]}(x). $$ Then $f$ is decreasing, $g(x)=f(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g(\mathrm e^{n^2})\geqslant\mathrm e^{-1}$ for every $n\geqslant1$, and $f\leqslant f_0$ hence $f$ is integrable.
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0Do you mean $f_0(x)=\sum_{n \ge 1}(n^2e^{n^2})^{-1}\mathbf{1}_{[0,e^{n^2}]}(x)?$ If so, why the integral of $f_0(x)$ is $\sum_{n \ge 1}n^{-2}$? – 2012-12-03
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0Because the integral of $c\mathbf 1_{[0,a]}$ with $a\gt 0$ is $ac$. – 2012-12-03
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0What motivated the definition of $f_0$? – 2012-12-03
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0The conditions have been changed. Is the answer correct now? – 2012-12-05
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0See my comment to the main post. – 2012-12-05
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0I am very sorry. I tried to post another question but the system told me that it is the same as the above one. I would like to post now the question with the additional condition, $f$ to be convex. Could anybody help me to post the new question? – 2012-12-07
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0@Did Why $g_0(e^{n^2})=1$? I think that $g_0(e^{n^2})>1.$ – 2018-03-08
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0Indeed, $g_0(\mathrm e^{n^2})\geqslant1$, not $g_0(\mathrm e^{n^2})=1$, thanks. (Would you be preparing yourself to *accept* an answer, 5+ years after the facts, by any chance? :-)) – 2018-03-08
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0The problem arised again these days and I remembered that I had the answer, but now I look at it more carefully. Thanks again. – 2018-03-08
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0Excellent! $ $ $ $ – 2018-03-08
I think it's not true, for example $f(x)=1/x$
and we have
$\lim_{x\rightarrow 0}xln(x)(1/x)=+\infty$
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2But $\int_0^\infty \frac 1 x$ does not exist IIRC – 2012-12-02
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0why do you check integral? – 2012-12-02
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5$f \in L^1(0,\infty)$ means that $\int_0^\infty f(x)dx < \infty$ – 2012-12-02
Edit: This argument assumes that the limit exists, which need not happen.
A change of variables gives, for $a>1$, $$ \int_{a}^{\infty} \frac{dx}{xln(x)} = \int_{\ln(a)}^{\infty} \frac{du}{u} =\infty $$ If the limit was not zero, call it $L$ (admitting $L=\infty$), we would get, for $x>a$ (with $a$ large enough) $f(x)\geq \frac{L}{2} \frac{1}{x\ln(x)}$ if $L$ is finite and $f(x)\geq \frac{1}{x\ln(x)}$ if it's infinite. Either way it contradicts the integrability of $f$ by the initial remark.
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2How do you know that the limit exists? The assumption would be that the $\limsup$ is not $0$. – 2012-12-02
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0Yeah, my mistake. For some reason I took $f$ to be increasing. I'll leave the answer though because the argument works if the limit is assumed to exist. – 2012-12-03