$L^p(\mu)$ space relations

Question

Let $(X,M, \mu)$ be a measure space. Suppose $1 \leq p < r \leq \infty$.

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Show that $f\in L^s(\mu)$ for $p \leq s \leq r$.

Attempt:

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Then $\int |f| ^p d\mu < \infty$ and $\int |f|^r d\mu < \infty$.

Isn't it obvious that $\int |f|^s d\mu < \infty$ already?

I'm bit confused on what to show.

Any help will be appreciated!

Thank you for your help!

Answer 1

This is Proposition 6.10 in Folland: If $0< p <q<r \leq \infty$, then $L^p \cap L^r \subset L^s$ and $\|f\|_s \leq \| f\|_p^{\lambda}\|f \|_r^{1- \lambda}$, where $\lambda \in (0,1)$ is defined by $\lambda = \frac{s^{-1} -r^{-1}}{p^{-1} - r^{-1}}$ when $r \ne \infty$, and $\lambda = p/s$ when $r \ne \infty$.

When $r= \infty$, you use the definition of $\| f\|_{\infty}$ to show the inequality.

When $r \ne \infty$, it's just Holder's inequality.

Answer 2

It seems obvious and it is to some degree but it requires a small amount of work to get there. Kavi's comment almost completely gives it away. If $|f(x)| \le 1$, then $|f(x)|^s \le |f(x)|^p$ since $p < s$. Similarly if $|f(x)| > 1$, then $|f(x)|^s \le |f(x)|^r$ since $s < r$. Can you put these together to show that $\int |f(x)|^s \,d\mu(x) < \infty$?

This is called ($L^p$) interpolation for what it's worth and is a very useful technique in functional analysis.

Answer 3

Can also be done from the Hölder inequality, in the form: If $1/\alpha+ 1/\beta = 1$, then

$$ \int |FG|\;d\mu \le \left(\int |F|^\alpha\;d\mu\right)^{1/\alpha}\;\left(\int |G|^\beta\;d\mu\right)^{1/\beta} \tag1$$

Given $p<s<r$ and $f \in L_p \cap L_r$ as in the problem, let $$ \alpha = \frac{r-p}{r-s},\quad \beta = \frac{r-p}{s-p} \quad\text{so that}\quad \frac{1}{\alpha}+\frac{1}{\beta} = 1\quad\text{and}\quad \frac{p}{\alpha} + \frac{r}{\beta} = s, $$ and let $$ F = |f|^{p/\alpha},\qquad G = |f|^{r/\beta} $$ Plug into $(1)$ to get $$ \int|f|^s\;d\mu \le \left(\int |f|^p\;d\mu\right)^{(r-s)/(r-p)} \;\left(\int |f|^r\;d\mu\right)^{(s-p)/(r-p)} < \infty. $$

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We will need a different (easier) argument in case $r = \infty$.