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Let $1 \leq p_1 \leq p_2 \leq +\infty$. Show that in a space of finite measure we have that $L^{p_2} \subset L^{p_1}$.

Could you give me some hints what I could do??

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Let $F = |f|^{p_1}$ and $G = 1$. Apply the Holder inequality

$||FG||_1 \leq ||F||_p ||G||_q$

where $p = p_2/p_1 > 1$ and $1/p + 1/q = 1$.

Note that $||G||_q = \mu(X)^{1-p_1/p_2}$ is finite as the underlying measure space $(X,\mu)$ is finite. This now gives you the bound you're looking for. Yes?


Added:

What we have so far is

$$\int |f|^{p_1} d\mu = \| FG \|_1 = \| |f|^{p_1}.1 \|_1 \ \leq \ \| \ \ |f|^{p_1} \ \|_{p_2/p_1} \ . \ \mu(X)^{1- p_2/p_1} \ \ \ \ \ --(*) $$

Now $$\| \ |f|^{p_1} \ \|_{p_2/p_1} = \left( \int \left(|f|^{p_1}\right)^{p_2/p_1} d\mu \right)^{p_1/p_2} = \left( \int |f|^{p_2} d\mu \right)^{p_1/p_2}$$

Substitute this expression back into (*) and take the $p_1$-th root of both sides,

$$ \left( \int |f|^{p_1} d\mu \right)^{1/p_1} \ \leq \ \left( \int |f|^{p_2} d\mu \right)^{1/p_2} . \left(\mu(X)^{1- p_2/p_1}\right)^{1/p_1}$$

That is

$$\| f \|_{p_1} \leq C \| f \|_{p_2}$$

where $C = \left(\mu(X)^{1- p_2/p_1}\right)^{1/p_1}$.

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  • $\begingroup$ Could you explain me why we take $p=p_2/p_1$?? Also, how did you find $||G||_q$?? $\endgroup$ – Mary Star Nov 10 '14 at 10:57
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    $\begingroup$ To show $L^{p_2} \subset L^{p_1}$ we want to show that $f \in L^{p_2} \rightarrow f \in L^{p_1}$. So if we can show that $$\| f \|_{p_1} \geq C \| f \|_{p_2}$$ for some constant $C$ we're done. The choice of $F$ and $G$ I've listed gives it to us. It's not obvious, but it works. $\|G\|_q$ is going to be just the measure of the space $X$ under $q = 1 - p_1/p_2$ because $$ \int_X 1^q \ d\mu = \mu(X) \rightarrow \|G\|_q = \left( \int_X 1^q \ d\mu \right)^{1/q} = \mu(X)^{1/q} = \mu(X)^{1 - p_1/p_2}$$ $\endgroup$ – Simon S Nov 10 '14 at 13:19
  • $\begingroup$ Could you explain me further why we have to show that $$||f||_{p_1}\geq C||f||_{p_2}$$ ?? $\endgroup$ – Mary Star Nov 10 '14 at 15:37
  • $\begingroup$ Ugh, sorry reversed the direction of the inequality there. We want to show $\|f\|_{p_2}$ bounded implies $\|f\|_{p_1}$ bounded. That is $$ \|f\|_{p_1} \leq C \|f\|_{p_2}$$ $\endgroup$ – Simon S Nov 10 '14 at 15:39
  • $\begingroup$ I got it!! What I stillndon`t understand is how we show the inequality $||f||_{p_1}\leq C ||f||_{p_2}$... Replacing F and G at Hilder s inequality we get $$|||f|^{p_1}||_{p_1}\leq |||f|^{p_1}||_{p_2/p_1} \mu(X)^{1-p_1/p_2}$$ how do we continue?? $\endgroup$ – Mary Star Nov 10 '14 at 15:59
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Apply Holder's inequality with conjugate exponents $p_{2}/p_{1}$ and $p_{2}/(p_{2}-p_{1})$. When $p_{2} < \infty$, this gives, for $f \in L^{p_{2}}$

$$\Vert f \Vert_{p_{1}}^{p_{1}} = \int |f|^{p_{1}} \cdot 1 \leq \Vert |f|^{p_{1}} \Vert_{p_{2}/p_{1}} \Vert 1 \Vert_{p_{2}/(p_{2}-p_{1})} = \Vert f \Vert_{p_{2}}^{p_{1}} \mu(X)^{p_{2}-p_{1}/p_{2}}$$

It's even easier to show when $p_{2} = \infty$.

Other than that, I'm not sure how you would show that the containments are strictly proper, though there are straightforward examples for $L^{1}$ and $L^{2}$.

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  • $\begingroup$ Could you explain why you take these conjugate exponents at Holder`s inequality?? Also you say for $f\in L^{p_2}$ and then you calculate the norm $p_1$, why?? $\endgroup$ – Mary Star Nov 10 '14 at 15:40

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