Weak convergence in $L^{∞}[0,1]$ implies strong convergence in $L^{p}[0,1]$ I'm trying to solve an exercise in Rudin's Functional Analysis. It is the problem 19 in Chapter 11.
The problem is:
If $f_n\in L^{∞}[0,1]$ and converges to $0$ in the weak topology of $L^{∞}[0,1]$, then $\int_0^1|f_n|^p\to 0$ for any $p\in (0,∞)$.
I have tried to use Gelfand Transform that Rudin mentioned in 11.13(f), which maps $L^{∞}[0,1]$ isometrically to $C(\Delta)$ as Banach Spaces, where $\Delta$ is the maximal ideal space. This approach seems to be hopeful because continuous functions are good, but I failed despite a few hours effort.
Another trial is to solve this just for p=2, but I did not proceed any more, too.
I think the conclusion of this problem is very powerful so I want to know how to solve it very much. So please tell me anything if you have an idea or some relative materials, thank you very much!
 A: Let me try $p=2$: Recall that the Gelfand Transform $f \mapsto \hat{f}$ is an isometric isomorphism onto $C(\Delta)$ and that there is a regular Borel measure $\mu$ on $\Delta$ with $\int_0^1 f dx = \int_\Delta \hat{f} d \mu$ $(f \in L^\infty[0,1])$. Setting $h(z)=|z|^2$ we have
$$
\forall \varphi \in \Delta: ~ |\hat{f}|^2(\varphi)= h\circ \hat{f}(\varphi)=h(\varphi(f))=\varphi(f)\overline{\varphi(f)}=\varphi(|f|^2)= \widehat{|f|^2}(\varphi),
$$
hence $|\hat{f}|^2=\widehat{|f|^2}$. Since $(f_n)$ is weakly convergent to $0$ it is bounded in $L^\infty([0,1])$, so $(\widehat{f_n})$ is pointwise convergent to $0$ in $\Delta$ and bounded in $C(\Delta)$. Hence $(\widehat{|f_n|^2})=(|\widehat{f_n}|^2)$ is pointwise convergent to $0$ and bounded. By LDCT
$$
\int_0^1 |f_n|^2 dx =\int_\Delta \widehat{|f_n|^2} d \mu =\int_\Delta |\widehat{f_n}|^2 d \mu \to 0 \quad (n\to \infty).
$$
If all above is correct (please check) the advantage of switching to $C(\Delta)$ is that (in contrast to $L^\infty[0,1]$) pointwise convergence to $0$ is easy to see.
Edit on account of Ryszard Szwarc' comment: With an analog calculation one gets $|\hat{f}|^{2m}=\widehat{|f|^{2m}}$ and $\int_0^1 |f_n|^{2m} dx \to 0$ for each $m \in \mathbb{N}$. If $p >0$ is given choose $m$ such that $2m \ge p$.
Then
$$
(\int_0^1 |f_n|^p dx)^{1/p} \le (\int_0^1 |f_n|^{2m} dx)^{1/(2m)} \to 0 \quad (n \to \infty).
$$
