I'm studying $L^p$ space.
$1 \le p < r then $L^p \cap L^q \subset L^r$. More over $L^p \cap L^\infty \subset L^r$
I'm trying to prove that fact. Which theorem is useful for proving that?
I'm studying $L^p$ space.
$1 \le p < r then $L^p \cap L^q \subset L^r$. More over $L^p \cap L^\infty \subset L^r$
I'm trying to prove that fact. Which theorem is useful for proving that?
Some hints:
$(1.)$ For $L^{p}\cap L^{q}\subset L^{r}$, take and $f\in L^{p}\cap L^{q}$ and divide the integration domain $X$ to $A:=\{x\in X:|f(x)|\leq 1\}$ and $A^{c}$. What can you say about $|f(x)|^{r}$ for $x\in A$ or $x\in A^{c}$? What can you conclude for $\|f\|_{r}$?
$(2.)$ Start showing that if $f\in L^{p}$ then $\mu(A^{c})<\infty$. Then use the fact that $|f(x)|\leq \|f\|_{\infty}$ for $\mu$-a.e. $x\in X$ and use the same logic as in the previous step to conclude $\|f\|_{r}<\infty$ if $f\in L^{p}\cap L^{\infty}$.
If you need some more hints or can't get started with these; we can discuss at the comment section below.
(1.) For every positive real number $a$, one has $a^r\leqslant a^p+a^q$ (one can prove separately the cases $a\leqslant1$ and $a\gt1$). Thus, $|f(x)|^r\leqslant |f(x)|^p+|f(x)|^q$ for every $x$.
Hence $\|f\|_r^r\leqslant\|f\|_p^p+\|f\|_q^q$ and $\|f\|_r$ is finite for every $f$ in $L^p\cap L^q$. Thus, $L^p\cap L^q\subset L^r$.
(2.) For every positive real numbers $a$ and $b$ such that $a\leqslant b$, one has $a^r\leqslant a^pb^{r-p}$. Thus, $|f(x)|^r\leqslant |f(x)|^p\cdot\|f\|_\infty^{r-p}$ for every $x$.
Hence $\|f\|_r^r\leqslant\|f\|_p^p\cdot\|f\|_\infty^{r-p}$ and $\|f\|_r$ is finite for every $f$ in $L^p\cap L^\infty$. Thus, $L^p\cap L^\infty\subset L^r$.
Note: Although such uniform pointwise inequalities cannot yield optimal norm inequalities, they are (i) simple to prove, and (ii) sufficient to get the inclusions of spaces the OP is interested in.
For $f(x)\ge0$, Jensen's Inequality yields $ \left(\frac{1}{\int_X f^p(x)\,\mathrm{d}x}\int_X f^{r-p}(x)f^p(x)\,\mathrm{d}x\right)^{\Large\frac{q-p}{r-p}}\le\frac{1}{\int_X f^p(x)\,\mathrm{d}x}\int_X f^{q-p}(x)f^p(x)\,\mathrm{d}x $ which becomes $ \left(\int_Xf^r(x)\,\mathrm{d}x\right)^{\Large\frac1r}\le\left(\int_Xf^p(x)\,\mathrm{d}x\right)^{\Large\frac1p\left(\frac{p}{r}\frac{q-r}{q-p}\right)}\left(\int_X f^q(x)\,\mathrm{d}x\right)^{\Large\frac1q\left(\frac{q}{r}\frac{r-p}{q-p}\right)} $ Thus for $f\in L^p\cap L^q$, $ \|f\|_r\le\|f\|_p^{\Large\frac{p}{r}\frac{q-r}{q-p}}\;\|f\|_q^{\Large\frac{q}{r}\frac{r-p}{q-p}}\tag{1} $ where $ \frac{p}{r}\frac{q-r}{q-p}+\frac{q}{r}\frac{r-p}{q-p}=1\tag{2} $ Note that when $q\to\infty$, $(1)$ becomes $ \|f\|_r\le\|f\|_p^{\Large\frac{p}{r}}\;\|f\|_\infty^{\Large1-\frac{p}{r}}\tag{3} $