Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p \neq 2$.

share|improve this question
    
I think that $f \in L^p(E)$ iff $p \geq 2$. –  Siméon Mar 18 '13 at 7:46

1 Answer 1

up vote 3 down vote accepted

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 $$\int_1^\infty f(x)^pdx \le \int_0^\infty e^{ay}dy = -1/a < \infty.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.