$L^1([0,1])$ closed unit ball is not weakly compact So I would like to prove this result by constructing a sequence of functions  $u_n$ in $L^1([0,1])$, such that $\|u_n\|_{L^1}\leq 1$ for all $n$, but this subsequence does not have a convergent subsequence. 
My idea was to take an approximation to the identity, say, $u_n = n\mathbf{1}_{[0,1/n]}$. Taking any continuous function on $\phi$ on $[0,1]$, ($\phi$ is also $L^{\infty}$ as a result), defines a linear functional on $L^1$:
$$
f(u_n) = \int_0^1 u_n\phi \mathrm{d}\mu
$$
And taking the limit to infinity, we see that $f(u_n) = \phi(0)$. Here is where my argument needs work: "This means that $u_n$ converges weakly to the Delta function, but this is not an element of $L^1$". How can I clarify this? Do I have to use the sequence:
$$
\int_0^1 u_n\phi \mathrm{d}\mu
$$
as a sequence in the dual of $C[0,1]$ and show then that $f_n(\phi) \xrightarrow{w*} \phi(0)$ ?
Thanks for the help.
 A: Here's an argument that shows that your sequence works:
Let $u_{n_k}$ be any subsequence of $u_n$. Assume that the weak limit of $u_{n_k}$ exists, denoted by $u$. Then, as $u_{n_k} \to 0$ almost everywhere, we have $u\equiv 0$, because weak limits and almost sure limits coincide (this can be seen by noting that both weak convergence and almost sure convergence cause convergence in probability of a subsequence). But if we let $\phi \equiv 1$, we find
$$ 1 = \int_{0}^{1} u_{n_k} \phi ~ \mathrm{d}\mu \to  \int_{0}^{1} u \phi ~ \mathrm{d}\mu = 0, $$
a contradiction.
A: Suppose that a subsequence $(u_{n_k})_{k\geqslant 1}$ converges weakly in $\mathbb L^1$ to some $u$. For all fixed $\delta$ and all Borel subset $B$ of $[0,1]$ included in $(\delta,1)$, using the definition of weak convergence with the linear functional associated to the indicator function of $B$, we derive that $\int_B u=0$ hence for all $N$, $u(x)=0$ for almost every $x\in (1/N,1)$. Therefore, the only possible limit for $(u_{n_k})_{k\geqslant 1}$ is $u=0$. But this does not work, for example using the functional $L\colon f\mapsto \int f$: $L(u_{n_k})=1$ but $L(u)=0$.
A: So, here is an answer that is inspired by Nate Eldrege's comments.
Let $u_n$ be an approximation to the identity. Clearly, $\|u_n\|_{L^1} \leq 1$ for all $n$. Assume now that $u_n$ has a weak subsequential limit, i.e $u_{n_k} \rightharpoonup u$ in $L^1$. Then, for any $L^\infty[0,1]$ function $\phi$ we must have:
$$
\int_0^1 u_{n_k}\phi \to \int_0^1 u \phi
$$
Moreover, we know by that $u_n$ has the property that:
$$
\int_0^1 u_n \phi \to \phi(0)
$$
for continuous $\phi$. Hence:
$$
\int_0^1 u_{n_k} \phi \to \phi(0) = \int_0^1 u\phi
$$
or every $\phi$ continuous. Now, we will show that no such $u$ in $L^1$ exists. Take the sequence $\phi_n = (1-x)^n$ on $[0,1]$ . Then, we see that:
$$
\int_0^1 u\phi_n =\phi_n(0) = 1
$$
for all $n$. It follows then that:
$$
\int_0^1 u\phi_n =\phi_n(0) \to  1
$$
By the dominated convergence theorem (bounding $u \phi_n$ by $u$, which we assume is $L^1$), we have that:
$$
\lim_{n \to \infty} \int_0^1 u\phi_{n} = \int_0^1 u\lim_{n \to \infty} \phi_n
$$
$\phi_n$ converges pointwise to the function $1$ at $0$, and $0$ elsewhere. Hence, we have:
$$
\lim_{n \to \infty} \int_0^1 u\phi_{n} = \int_0^1 u\lim_{n \to \infty} \phi_n = 0
$$
But we showed above that the limit is $1$. So this cannot happen.
