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

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    xln(x)f(x) $\qquad$ or $\qquad$ xln(xf(x))??2012-12-02
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    $f(x) x \ln x $2012-12-02
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    The 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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    @DavideGiraudo Good analysis. There is.2012-12-02
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    Yes, for example $t_k=2^{k^2}$.2012-12-03
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    OP: 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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    I 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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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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    Do 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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    Because the integral of $c\mathbf 1_{[0,a]}$ with $a\gt 0$ is $ac$.2012-12-03
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    What motivated the definition of $f_0$?2012-12-03
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    The conditions have been changed. Is the answer correct now?2012-12-05
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    See my comment to the main post.2012-12-05
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    I 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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    @Did Why $g_0(e^{n^2})=1$? I think that $g_0(e^{n^2})>1.$2018-03-08
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    Indeed, $g_0(\mathrm e^{n^2})\geqslant1$, 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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    The 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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    Excellent! $ $ $ $2018-03-08
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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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    But $\int_0^\infty \frac 1 x$ does not exist IIRC2012-12-02
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    why do you check integral?2012-12-02
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    $f \in L^1(0,\infty)$ means that $\int_0^\infty f(x)dx < \infty$2012-12-02
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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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    How do you know that the limit exists? The assumption would be that the $\limsup$ is not $0$.2012-12-02
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    Yeah, 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