Lebesgue Integration problem Can I ask a homework question here?
Assume that $f \in L^q(\mathbb R^d)$ for some $ q < \infty$ . show that
$\mathrm{lim}_{p \to \infty}||f||_p = ||f||_{\infty}$
$p$ conjugate of $q$
 A: In order your question to be correct you need to require that for some $q$ we have $f\in L^p(\mathbb{R}^d)$ for all $p\geq q$. To solve this problem use the following approach
1) Take $\{p_n:n\in\mathbb{N}\}$ such that $\lim\limits_{n\to\infty} p_n=+\infty$ and $\lim\limits_{n\to\infty}\Vert f\Vert_{p_n}=\liminf\limits_{p\to\infty}\Vert f\Vert_p:=a$.
2) Using Chebyshev inequality prove that $\lim\limits_{n\to\infty}\mu(\{x\in\mathbb{R}^d:|f(x)|> a+\varepsilon\})=0$ for all $\varepsilon>0$.
3) Conclude that $\Vert f\Vert_\infty\leq a$.
4) Prove that $\Vert f\Vert_p\leq \Vert f\Vert_q^{q/p}\Vert f\Vert_\infty^{1-q/p}$ for all $p> q$ and then conclude $\limsup\limits_{p\to\infty}\Vert f\Vert_p\leq\Vert f\Vert_\infty$.
5) The rest is clear.
A: Let $\mu:=|f|^qd\lambda$. 


*

*Check that $f\in L^{\infty}(\lambda)$ if and only if $f\in L^{\infty}(\mu)$. 

*Show that $\lVert f\rVert_{L^p(\lambda)}=\lVert f\rVert_{L^{p-q}(\mu)}^{\frac{p-q}q}$.

*Use this to show the result when $f\in L^{\infty}$. 

*When $f\notin L^{\infty}$, use the sets $B_n:=\{x, |f(x)|\geq n\}$.  

A: This solves the question under the hypothesis that $\|f\|_q$ is finite for some fixed positive finite $q$.
Assume without loss of generality that $\|f\|_\infty=1$. Then:


*

*$\|f\|_p^p=\int |f|^p\leqslant\int |f|^q=\|f\|_q^q$, hence $\|f\|_p\leqslant\|f\|_q^{q/p}$, where $\|f\|_q^{q/p}\to1$ when $p\to\infty$.

*In the other direction, by definition of $\|f\|_\infty=1$, for every $u\lt1$ there exists a measurable set $A$ of positive finite Lebesgue measure $m$ such that $|f|\geqslant u$ uniformly on $A$. Thus, $\|f\|_p^p\geqslant\int_A u^p=mu^p$ and $\|f\|_p\geqslant m^{1/p}u$, where $m^{1/p}u\to u$ when $p\to\infty$.


Thus, for every $u\lt1$, $u\leqslant\liminf\limits_{p\to\infty}\|f\|_p\leqslant\limsup\limits_{p\to\infty}\|f\|_p\leqslant1$, hence $\lim\limits_{p\to\infty}\|f\|_p=1$.
Now, show that the without loss of generality step at the beginning of the post does not make all the rest a proof by intimidation.
