Finding *-weak limit of functional I´m trying to prove that the following functional sequence converges *-weakly, finding it´s limits:
$$\lambda_n(x)=\int_{-1}^{1}x(t)\cos(n\pi t)dt,\;\;x\in L^2(-1,1),\;\; n\in\mathbb{N}$$
My idea is to assume that there is such a functional, i.e exists $\lambda\in L^2(-1,1)^*$, such that 
$$\lim_{n\to\infty}\lambda_n(x)=\lambda(x),\;\;\forall x\in L^2(-1,1)$$
and then proceed in this way:
let $\epsilon>0$. Then, exists $N\in\mathbb{N}$, such that for all $n>N$, we have
$$
|\lambda_n(x)-\lambda(x)|<\epsilon\;\;\;\;\;(1)$$
Now i have a guess, of the form of this $\lambda$, but i´m not really sure.
$$\lambda(x)=\int_{-1}^{-1}x(t)g(t)dt$$
Asuming this true, we would have from (1) and from C-S inequality, that
$$\Big|\int_{-1}^{-1}x(t)(\cos(n\pi t)-g(t))dt\Big|\leq \Bigg[\int_{-1}^{1}x^2(t)dt\Bigg]^\frac{1}{2}\Bigg[\int_{-1}^{1}(\cos(n\pi t)-g(t))^2dt\Bigg]^\frac{1}{2}$$
Since $x\in L^2(-1,1)$, its integral is finite, and then the only thing left to do is to find the $g\in L^2(-1,1)$ such that makes the right expresion in the inequality as "little as i want". Maybe $\lim_{n\to\infty} \cos(n\pi t)=g(t)$ But i´m not sure.
Is this procedure well? and if so, how can i find such $g$?. Thank you
 A: We show that the sequence converges in the weak* topology to $0$. To do this, we need to prove that
$$
\int_{-1}^1 x(t)\cos(n\pi t)\,dt \to 0
$$
as $n\to\infty$ for every $x\in L^2(-1,1)$.
In fact, since $L^2(-1,1)\subseteq L^1(-1,1)$, it suffices to prove the above for every $x\in L^1(-1,1)$.
Step 1: Suppose $x=1_{[a,b]}$ is the indicator function of some interval $[a,b]\subseteq (-1,1)$. Then
$$
\int_{-1}^1x(t)\cos(n\pi t)\,dt
= \int_a^b\cos(n\pi t)\,dt
= \frac{1}{n\pi}\left(\sin(n\pi b)-\sin(n\pi a)\right)
\to 0.
$$
Step 2: If $x=\sum_{i=1}^nc_i1_{[a_i,b_i]}$ is a step function, then
$$
\int_{-1}^1x(t)\cos(n\pi t)\,dt
= \sum_{i=1}^nc_i\int_{a_i}^{b_i}\cos(n\pi t)\,dt \to 0
$$
by the previous step.
Step 3: Given $x\in L^1(-1,1)$ and $\varepsilon>0$, we can use the density of step functions to find a step function $h$ such that $\int_{-1}^1|x(t)-h(t)|\,dt<\varepsilon/2$. By the previous step, there is $N\in\mathbb{N}$ such that $|\int_{-1}^1h(t)\cos(n\pi t)\,dt|<\varepsilon/2$ for $n\ge N$.
Then, if $n\ge N$, we have
\begin{align*}
\left|\int_{-1}^1x(t)\cos(n\pi t)\,dt\right|
&\le \int_{-1}^1|\cos(n\pi t)||x(t)-h(t)|\,dt + \left|\int_{-1}^1h(t)\cos(n\pi t)\,dt\right| \\
&< \varepsilon.
\end{align*}
Therefore $\lambda_n\to 0$ in the weak* topology.
