Show inclusion of $L^p$ spaces in a space of finite measure

Question

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??

Answer 1

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?

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

Answer 2

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 $L^{2}$.