Prove that the unit ball in $L^{p}([0,1]),$ where $1 \leq p \leq \infty,$ is not strongly compact. Give an example of a bounded sequence in $L^{1}([0,1])$ that does not have a weakly convergent subsequence. Why doesn't this contradict the Banach Alaoglu theorem?
I was not able to progress because I'm not able to understand strong compactness.
Just for convinience, let me consider $L^1([-1,1])$. Take $f_n(x) = 0$ if $\lvert x \rvert >1/2n$, otherwise put $f_n(x) = n$. It is clear that $\int_{[-1,1]} \lvert f_n(x) \rvert \ dx = 1 \ \forall n$, and hence the sequence is bounded. Now remember that the dual space of $L^1$ is $L^{\infty}$ in the sense that if $T \in (L^1)^*$ then there is a function $\varphi \in L^{\infty}$ such that
$$T(f) = \int_{[-1,1]} f(t)\varphi(t) \ dt$$
Considering this, a sequence $f_n$ is weakly convergent iff there is a function $f \in L^1$ such that
$$\int_{[-1,1]} f_n(t)\varphi(t) \ dt \to \int_{[-1,1]} f(t)\varphi(t) \ dt \ \ \ \ \forall \varphi \in L^{\infty}$$
But this can't happen with any subsequence of $f_n$. For this observe that if $\varphi$ is continuous then
$$\int_{[-1,1]} f_n(t)\varphi(t) \ dt \to \varphi(0)$$
and clearly $\varphi(0)$ can't be written as $\int_{[-1,1]} f(t)\varphi(t) \ dt$ for any $f$, because we can modify our $\varphi$ changing the value of the integral but not the value of $\varphi(0)$.
To see why $\int_{[-1,1]} f_n(t)\varphi(t) \ dt \to \varphi(0)$
$$\Biggl\lvert \varphi(0) - \int_{[-1,1]} f_n(t)\varphi(t) \ dt \Biggr\rvert = \Biggl\lvert \int_{[-1,1]} f_n(t)(\varphi(0)-\varphi(t)) \ dt \Biggr\rvert = \Biggl\lvert \int_{-1/2n}^{1/2n} n(\varphi(0)-\varphi(t)) \ dt \Biggr\rvert $$
since $\varphi$ is continuous $\lvert \varphi(t)-\varphi(0)\rvert \leq \epsilon$ if $t$ belongs to $[-1/2n,1/2n]$ for $n$ great enough. Then, applying triangular inequality
$$\Biggl\lvert \int_{-1/2n}^{1/2n} n(\varphi(0)-\varphi(t)) \ dt \Biggr\rvert \leq \epsilon \int_{-1/2n}^{1/2n} n\ dt = \epsilon$$
This example doesn't contradict the Banach Alaoglú theorem since this theorem applies to $X^*$. What happens is that $L^1$ is not the dual space of any $X$.
With respect to the question about the unit ball of $L^p$, the answer is given here en.wikipedia.org/wiki/Riesz%27s_lemma, where the Riesz lemma is used. This lemma tells you that in a banach space the unit ball is compact iff the space has finite dimension.