Suppose $\mu(X)<\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q<\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r<q$

Question

Suppose $\mu(X)<\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q<\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r<q$

In the solution the assistant wrote this:

If $1\le r<q$ then $q/r\ge 1$. Let p be the dual exponent to $q/r$ (i.e. $\frac{1}{q/r}+\frac{1}{p}=1)$

then by Hölder Inequality

$\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}<\infty$

for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ?

instead of this can't we say (only in this case) that

$\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r$ then

$(\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r<\infty$ ?

Answer 1

A direct approach: since $r\leqslant q$, $|f|^r\leqslant|f|^q+1$ pointwise, and $1$ is integrable because $\mu$ has finite mass.

Note: The proof in your post is Hölder's inequality, applied to the functions $u=|f|^r$ and $v=1$, and to the conjugate exponents $q/r$ and $p$. Then $|u|^{q/r}=|f|^q$ and $|v|^p=1$.