0
$\begingroup$

Assume we have $1\leq p<q<\infty$. How can I show that $L_p(\mathbb{R})\neq L_q(\mathbb{R})$?

I suppose the easiest way would be to show that neither is a subset of the other, but how would I get started on that?

$\endgroup$
0
$\begingroup$

Note $L^q(\mathbb{R})\subset L^p(\mathbb{R})$. Let $$ f(x)=\left\{\begin{array}{ll} \frac{1}{x^\alpha}&\text{ if }x\in(0,1)\\0&\text{ else}\end{array}\right. $$ where $\alpha=\frac{p+q}{2pq}$. Clearly $$ \int_{\mathbb{R}}|f|^pdx=\int_0^1\frac{1}{x^\frac{p+q}{2q}}dx<\infty$$ and $$ \int_{\mathbb{R}}|f|^qdx=\int_0^1\frac{1}{x^\frac{p+q}{2p}}dx=\infty. $$ So $L^q(\mathbb{R})\not=L^p(\mathbb{R})$.

$\endgroup$

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.