Htykuut

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, $u_n = nmathbf{1}_{[0,1/n]}$. Taking any continuous function on $phi$ on $[0,1]$, ($phi$ is also $L^{infty}$ as a result), defines a linear functional on $L^1$:
$$
f(u_n) = int_0^1 u_nphi mathrm{d}mu
$$
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:
$$
int_0^1 u_nphi mathrm{d}mu
$$
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.

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 $uequiv 0$, 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.

Suppose that a subsequence $(u_{n_k})_{kgeqslant 1}$ 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 $xin (1/N,1)$. Therefore, the only possible limit for $(u_{n_k})_{kgeqslant 1}$ is $u=0$. But this does not work, for example using the functional $Lcolon fmapsto int f$: $L(u_{n_k})=1$ but $L(u)=0$.

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:
$$
int_0^1 u_{n_k} phi to phi(0) = int_0^1 uphi
$$
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:
$$
int_0^1 uphi_n =phi_n(0) = 1
$$
for all $n$. It follows then that:
$$
int_0^1 uphi_n =phi_n(0) to 1
$$
By the dominated convergence theorem (bounding $u phi_n$ by $u$, which we assume is $L^1$), we have that:
$$
lim_{n to infty} int_0^1 uphi_{n} = int_0^1 ulim_{n to infty} phi_n
$$
$phi_n$ converges pointwise to the function $1$ at $0$, and $0$ elsewhere. Hence, we have:
$$
lim_{n to infty} int_0^1 uphi_{n} = int_0^1 ulim_{n to infty} phi_n = 0
$$
But we showed above that the limit is $1$. So this cannot happen.