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I'm confused about the proof for this theorem: let $E$ be a measurable set s.t. $mE<\infty$ and $1 \leq p_1 < p_2 \leq \infty$. Then, $L^{p_2}(E) \subset L^{p_1}(E)$. Also, $\|f\|_{p_1} \leq c\|f\|_{p_2} \forall f \in L^{p_2}$, where $c=[mE]^{\frac{p_2-p_1}{p_1p_2}}$ when $p_2<\infty$.

I'm trying to thoroughly comprehend the proof; here's what I understand so far:

Consider the case when $p_2 < \infty$. Let $p=\frac{p_2}{p_1}>1$. Let $q$ be the conjugate of $p$. By assumption, $f \in L^{p_2}$. This is silly, but the part I'm confused on is, after this, we have $f^{p_1} \in L^p$. How?

Bhaskar Vashishth
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2 Answers2

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For the case $1\leq p_1 < p_2 < \infty$, let $p=p_2/p_1 >1$. Then take any $f \in L^{p_2}(E)$, this means $$ \int_{E} |f|^{p_2} dm < \infty $$ Now, using that $p=p_2/p_1$ we observe that $$ \int_{E} |f^{p_1}|^{p} dm = \int_{E} |f|^{ p_1 \cdot p} dm =\int_{E} |f|^{p_2} dm < \infty $$ Thus indeed $f^{p_1} \in L^p(E)$. Now for the rest, note that if $q$ is the conjugate of $p$ ($1/p + 1/q=1$), then clearly, since $m(E)<\infty$, the constant function $1 \in L^q(E)$, and then by Hölder inequality we get that $$ \int_E |f|^{p_1}dm=\int_E |f|^{p_1} \cdot 1 dm \leq \left(\int_E \left| \ |f|^{p_1} \right|^p dm \right)^{1/p} \cdot \left(\int_E 1^q dm\right)^{1/q} = \left(\int_E |f|^{p_2} dm \right)^{1/p} \cdot (m(E))^{1/q} $$ But the last part is $< \infty$ since $\int_E |f|^{p_2} dm<\infty$ and $m(E)<\infty$. So we just get that $\int_E |f|^{p_1}dm< \infty$, and hence $f\in L^{p_1}(E)$, which implies indeed that $L^{p_2}(E) \subset L^{p_1}(E)$.

Finally, since $1/p+1/q=1$, then $q=p/(p-1) = p_2/(p_2-p_1)$. Then the last inequality can be seen as $$ \| f \|^{p_1}_{p_1} \leq \left(\| f \|_{p_2}^{p_2}\right)^{1/p} (m(E))^{\frac{p_2-p_1}{p_2}} = \| f \|_{p_2}^{p_1} (m(E))^{\frac{p_2-p_1}{p_2}} $$ Then, elevating to $1/p_1$ both sides, we conclude that indeed $$ \| f \|^{p_1} \leq \| f \|_{p_2} (m(E))^{\frac{p_2-p_1}{p_1p_2}} $$

Alonso Delfín
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Given $f \in L^{p_2}$, $f^{p_2} \in L^1$. So since $p_2 = p_1p$, $(f^{p_1})^p\in L^1$, i.e., $f^{p_1}\in L^p$.

kobe
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