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For $E = (1, \infty)$ and $f$ defined by $$f(x) = \frac{x^{-1/2}}{1 + \ln{x}} \text{for } x >1$$ $f$ belongs to $L^{p}{(E)}$ iff $p = 2$

This is an example on page 143 of Real Analysis, Royden et al(4ed).

I know that $f \in L^2(E)$, but I've no idea how to show $f \notin L^{p}(E)$, for $p \neq 2$.

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  • $\begingroup$ I think that $f \in L^p(E)$ iff $p \geq 2$. $\endgroup$
    – Siméon
    Mar 18, 2013 at 7:46

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Substitution $x=e^y$ makes $\int_1^\infty f(x)^p dx = \int_0^\infty exp((1-p/2)y)/(1+y)^p dy$. For $p<2$ the integrand diverges to $\infty$ and for $p>2$ the numerator is already integrable.

More explicitly, for $p>2$ we get with $a=(1-p/s)<0$ that $$\int_1^\infty f(x)^pdx \le \int_0^\infty e^{ay}dy = -1/a < \infty.$$

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