Let $(X, \Sigma, \mu)$ be a finite measure space. Show that if $1 \leq p < r < \infty$ then there is a finite constant $c$ such that $||u||_p \leq c||u||_r$ for every $u \in \ L^r(μ)$. Find a condition on $\mu$ for which $c=1$.
My answer so far:
We want to show: $\displaystyle \ (\int |u|^p d\mu)^{1/p} \le C(\int |u|^r dμ)^{1/r} $
Hence we want to show: $\displaystyle \int |u|^p dμ \le C^p(\int |u|^r dμ)^{p/r} $ (Taking the p-th power of both sides)
Using Holders Inequality on the functions $|u|^p$ and $1$ we have:
$\displaystyle \int |u|^p d\mu \leq (\int |u^p|^{r/p} d\mu)^{p/r} (\int 1 d\mu)^{(r-p)/r} = C^p(\int |u|^r d\mu)^{p/r}$ as $p/r + (r-p)/r$ must equal $1$
Hence $\displaystyle c = (\int 1 d\mu)^{(r-p)/pr} = \mu(X)^{(r-p)/pr} $
This is where I'm stuck as the answer says $c = \mu(X)^{r/(r-p)} $
Edit: Found a different exam paper with the same question confirming the solution is $\displaystyle c = (\int 1 d\mu)^{(r-p)/pr} = \mu(X)^{(r-p)/pr} $
You need to think of $1$ as the indicator function of the whole space.
So $\int 1 d\mu = \int 1_X d\mu=\mu(X)$
Taking pth powers of your statement we get:
$||u||_p= \displaystyle \left( \int |u|^p d\mu \right)^{1/p} \leq \left(\int |u^p|^{r/p} d\mu \right)^{1/r} \left(\int 1 d\mu \right)^{(r-p)/pr} = c \left(\int |u|^r d\mu \right)^{1/r}=c||u||_r$
So $\displaystyle c=\left(\int 1 d\mu \right)^{(r-p)/pr}=\mu(X)^{(r-p)/pr}$
I think the answer supplied is wrong.