Showing that $x_{n}$ does not converge weakly to $0$ in $L^{\infty}([0,1])$

72 Views Asked by At

Background:

I want to show that when $ p= \infty$, the following are not equivalent for $(x_{n})_{n} \subseteq L^{\infty}([0,1])$:

$1.$ $x_{n} \xrightarrow{ w} 0$

$2.$ $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}<\infty $ and $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$.

As a hint, consider $x_{n}(t):=t^{n}$

It is clear that in this case $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{\infty}\leq 1<\infty$ and also that $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$

so it needs to fail at $1.$: But which functional $\ell \in (L^{\infty}[0,1])^{*}$ can I actually use?

1

There are 1 best solutions below

0
On BEST ANSWER

This is not correct. The sequence $x_n(t)=t^{n}$ does not work. If $g \in (L^{1}[0,1])^{*}=L^{\infty}[0,1]$ then $\int_0^{1} t^{n}g(t)\, dt \to 0$ by DCT so 1) is true in this case.

It appears that the question is supposed to be about $(x_n)$ as a sequence in $L^{\infty}$. To show that 1) fails in this case consider the maps $f \to f(1)$ in $C[0,1]$. Extend it to $L^{\infty}$ by Hahn Banach Theorem. This gives you the functional $\ell$ you are looking for.