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!

  • $\begingroup$ 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

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 $p_0 < p \le p_1$.

  • $\begingroup$ 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

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