I tried to solve the following problem and want to know if my approach is correct and some help in order to finish. Thanks.
Let $V = L^\infty(\mathbb{R})$ equipped with the norm $\left\lVert \cdot\right\rVert_\infty$ and $V^*$ its dual space.
For any $f\in L^1(\mathbb{R})$, define $l_f :V\to \mathbb{R}$ by $$l_f(g) = \int f\cdot g \,dx \,\,\,\,\,\,\, \forall g\in V$$
- Show that $l_f\in V^*$.
Proof:
Let $g,h\in V$ and $\lambda,\mu\in\mathbb{R}$. Then $$l_f(\lambda g+\mu h) = \int f\cdot (\lambda g +\mu h)\,dx = \int \lambda\, f\cdot g\,dx\,+\,\int \mu\, f\cdot h\,dx=\lambda\,l_f(g)\,+\,\mu\, l_f(h)$$ Hence $l_f$ is linear.
We now look at $\left\lVert l_f \right\rVert_{op}$. By definition,
$$\left\lVert l_f \right\rVert_{op}=\sup\big\{\lvert l_f(g)\rvert : \left\lVert g \right\rVert_{\infty}\leq 1\big\}$$
Then we notice that
$$\bigg\lvert\int f\cdot g\,dx\bigg\rvert\leq \int\lvert f\cdot g\rvert \,dx\leq \left\lVert g\right\rVert_{\infty}\int\lvert f\rvert\,dx = \left\lVert g\right\rVert_{\infty}\left\lVert f\right\rVert_{1}$$
It follows that, when $\left\lVert g\right\rVert_{\infty}, \leq 1$, we have $$\left\lVert l_f \right\rVert_{op}\leq\left\lVert f\right\rVert_{1}<\infty$$
Hence, $l_f$ is bounded and it belongs to $V^*$. $\square$
- For each $n\in\mathbb{N}$, define $l_n:V\to\mathbb{R}$ by
$$l_n(g) =\frac{1}{2n}\int_{-n}^n g\,dx$$
Show that there exists an $l\in V^*$ and a subsequence $(n_k)_{k\in\mathbb{N}}$ such that $l_{n_k}\xrightarrow{\text{w*}} l$.
Proof:
First notice that $\left\lVert l_n \right\rVert_{op}\leq 1$, indeed, by definition $$\left\lVert l_n \right\rVert_{op} = \sup\big\{ \lvert l_n(g)\rvert : \left\lVert g \right\rVert_{\infty}\leq 1\big\}$$
Now
$$\bigg\lvert \frac{1}{2n}\int_{-n}^n g\,dx \bigg\rvert \leq \frac{1}{2n} \int_{-n}^n\lvert g\rvert\,dx \leq \left\lVert g \right\rVert_{\infty} \frac{1}{2n} (n-(-n)) =\left\lVert g \right\rVert_{\infty}$$
By above, for all $n\in\mathbb{N}$, we have $l_n \in \overline{B} {}^* = \bigg\{f\in V^* : \left\lVert f \right\rVert_{op}\leq 1\bigg\}$.
By the Banach-Alaoglu Theorem, we know that $\overline{B} {}^*$ is weak* compact. And hence $l_n$ shoud have a convergent subsequence in the weak* topology.
However, I don't know how to define this $l\in V^*$.
