# $L^1$ is not reflexive

I want to show that $L^1(\mathbb{R}^n)$ is not reflexive. So we look at a sequence of functions $(\delta_k)\in L^1$ with $\delta_k:\mathbb{R}^n\rightarrow\mathbb{R}$ and

$1)\ \delta_k\geq 0 \ \forall k\in\mathbb{N}$,

$2)\ \int_{\mathbb{R}^n}\delta_k=1 \ \forall k\in\mathbb{N}$

$3)\ \int_{\mathbb{R}^n\backslash (B_{\frac{1}{m}}(0))}\delta_k\xrightarrow{k\rightarrow\infty}0 \ \forall m\in \mathbb{N}$

This is a Dirac-sequence. Now I want to show that there is such a sequence in $L^1$ and that there is no weakly convergent subsequence. With that I have to conclude that $L^1$ is not reflexive.

I already showed the existence of the dirac-sequence. But I don't know why there is no subsequence. Can someone also eplain why we can conclude that $L^1$ is not reflexive?

• You can use Radon-Nykodim theorem to show that such a function cannot exist in $L_1(\mu)$ since it does not induce a measure $\nu:\nu<<\mu$ – Gabriele Cassese Jan 25 '18 at 23:13
• Another way to see this, in a reflexive space the unit ball is weakly compact. Now, the Eberlein–Šmulian theorem says that a set in a Banach space is weakly compact if and only if is sequentially weakly compact. – clark Jan 25 '18 at 23:20
• In my opinion, the easiest strategy to see that $L^1$ is not reflexive is to check that $L^1$ is separable while $(L^1)^* = L^\infty$ is not separable. – gerw Jan 26 '18 at 7:21

Hint Assume by contradiction that this sequence has a convergent subsequence $\delta_{k_n} \to f$.
Now, for each $a >0$, $L_a(g) = \int_{\mathbb{R}^n\backslash (B_{a}(0))} g(t) dt$ is a linear functional on $L^1(\mathbb R^n)$ and $\delta_{k_n} \to f$ weakly, hence
$$L_a(\delta_{k_n}) \to L_a(f)$$
Also, $T(g)= \int_{\mathbb{R}^n} g(t) dt$ is a linear functional on $L^1(\mathbb R^n)$ and hence $$T(\delta_{k_n}) \to T(f)$$
It is easy to show that the value you get for $L_a(f) \forall a>0$ and $T(f)$ are not consistent, i.e. contradiction.