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We consider Lebesgue measure space $(\mathbb{R} , \mathcal{L} , m)$ and let $f : \mathbb{R} \to \mathbb{R}$, $f(x) = \frac{1}{x^{\frac{1}{2}} (1 + |\log x|)} {\chi}_{(0 , \infty)}(x)$ for all $x \in \mathbb{R}$. Show that $f \in L^p$ if, and only if, $p = 2$. Obviously, $f \geq 0$. I have already shown that $f \in L^2$ with ${\|f\|}_2 = \sqrt{2}$ but I can't see that $f \notin L^p$, $p \neq 2$. Any help? Thank you very much.

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    For $p > 2$, look at the behaviour on $(0,1)$. For $p < 2$, look at $(1,+\infty)$. – 2017-01-10
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    Yes the problem would finish if I could find $g : (1 , \infty) \to \mathbb{R}$, with $g \leq f$ on $(1 , \infty)$, such that $g \notin L^p$, $p < 2$, and $h : (0 , 1) \to \mathbb{R}$, with $h \leq f$ on $(0 , 1)$, such that $h \notin L^p$, $p > 2$. I have looked for but I didn't find $g$ and $h$. – 2017-01-10
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    Compare $f(x)^p$ to $x^{-1}$. – 2017-01-10
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    I begin with the case $p > 2$. How can I prove that exists $y \in (0 , 1)$ such that $1/x \leq f^p(x)$ for all $x \in (0 , y)$? – 2017-01-10
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    $$\frac{1}{x f(x)^p} = \frac{x^{p/2}(1+\lvert\log x\rvert)^p}{x} = x^{p/2 - 1}(1+\lvert\log x\rvert)^p$$ – 2017-01-10
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    Sorry I can't see the argument. What must I do with the expression? – 2017-01-10
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    Showing that its (right hand side) limit at $0$ is $0$ would suffice. – 2017-01-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51525/discussion-between-joseabp91-and-daniel-fischer). – 2017-01-10
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    I think that I have solved the problem. Thank you Daniel. Like+1 – 2017-01-10

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$\int_1^{\infty} \frac{1}{x^{\frac{p}{2}}(1+log(x))^p}dx = \int_0^{\infty} \frac{1}{e^{x\frac{p}{2}}(1+x)^p}e^xdx = \int_0^{\infty} e^{x(1-\frac{p}{2})}(1+x)^{-p} dx$

What is inside the last integral diverges for any $p<2$.

and $\int_0^1\frac{1}{x^{\frac{p}{2}}(1-log(x))^p}dx= \int_{-\infty}^0 e^{x(1-\frac{p}{2})}(1-x)^{-p} dx$ and what is in the last integral diverges for any $p>2$ as $x \to -\infty$