Suppose $1\le p < r < q < \infty$. Prove that $L^p\cap L^q \subset L^r$.

Suppose $1\le p < r < q < \infty$. Prove that $L^p\cap L^q \subset L^r$.

So suppose $f\in L^p\cap L^q$. Then both $\int |f|^p d\mu$ and $\int|f|^q d\mu$ exist. For each $x$ in the domain of $f$, $|f(x)|^r$ is between $|f(x)|^p$ and $|f(x)|^q$.

So $|f|^r\le\max(|f|^p,|f|^q)=\frac12({|f|^p+|f|^q+||f|^p-|f|^q|})$.

The RHS is integrable, and since $|f|^r$ is bounded above by an integrable function, it is also integrable. So $f\in L^r$.

Is my argument correct?

Actually is this still true when $q=\infty$? In this case, we have $1\le p < r < \infty$. Is $L^p\cap L^{\infty}$ necessarily a subset of $L^r$? Why?

Please show that if $f\in L^p\cap L^{\infty}$, then $||f||_r \le ||f|_p^{p/r} ||f||_{\infty}^{1-p/r}$

EDIT: Thanks for the link to the first part of my question. I'm still quite interested in the case $q = \infty$.

Solution 1:

From Lyapunov's inequality we have $$ \Vert f\Vert_r\leq\Vert f\Vert_p^{\frac{p}{r}\frac{r-q}{p-q}}\Vert f\Vert_q^{\frac{q}{r}\frac{p-r}{p-q}}\tag{1} $$ So we proved that $L_p\cap L_q\subset L_r$. To handle the case $q=\infty$ note that $$ \Vert f\Vert_r =\left(\int_X |f(x)|^{r-p}|f(x)|^p dx\right)^{1/r} \leq\left(\int_X |f(x)|^p dx\right)^{1/r}\left(\operatorname{esssup}\limits_{x\in X}|f(x)|^{r-p}\right)^{1/r}\\ =\Vert f\Vert_p^{p/r}\Vert f\Vert_\infty^{1-p/r} $$ Which shows that $L_p\cap L_\infty\subset L_r$.