4
$\begingroup$

Suppose $0 < p_0 < p_1 \leq \infty$. Find examples of functions $f$ on $(0,\infty)$ with Lebesgue measure such that $f \in L^p$

if and only if

(a) $p_0 < p < p_1$ (b) $p_0 \leq p \leq p_1$ (c) $p=p_0$

The hint says consider functions of the form $f(x) = x^{-a}|\log x|^b$ but the hint does not help since integral is divergent for any choice of $a$ and $b$

Thank you for your time!

$\endgroup$
  • $\begingroup$ Yes, that is if your $f$ is over the entire interval. That does not have to the be the case. You can consider $x\log x$ on $(1,\infty)$, say... $\endgroup$ – Lost1 Mar 20 '13 at 12:12
4
$\begingroup$

I'll give you one example and let you fill in the details and find the other examples.


This integral converges iff $p > p_0$: $$ \int_2^\infty \frac{dx}{x^{p/p_0}} $$

You can show this by direct computation.


This integral converges iff $p \le p_1$: $$ \int_0^{1/e} \frac{dx}{\left(x(\log x)^2\right)^{p/p_1}} $$

The case $p = p_1$ follows by direct computation. The cases $p < p_1$ and $p > p_1$ follow by comparison with integrals similar to the previous intergral.


Put: $$ f(x) = \frac{\chi_{(2,\infty)}}{x^{1/p_0}} + \frac{\chi_{(0, 1/e)}}{\left(x(\log x)^2\right)^{1/p_1}} $$

And show that $f \in L^p((0, \infty))$ iff $p_0 < p \le p_1$.

$\endgroup$
  • $\begingroup$ How are you supposed to evaluate this integral$$ \int_0^{1/e} \frac{dx}{\left(x(\log x)^2\right)^{p/p_1}} $$ $\endgroup$ – BronchoX Nov 29 '16 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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