limit of a sequence of functionals Consider a sequence of functionals $(f_n)$, $f_n(x)=\int_{-1}^1x(t)\cos(nt)dt,\ n\geq 1$,  on the space $L_2(-1,1)$. I need to prove that $f_n(x)\to 0$, as $n\to\infty$, for all $x\in L_2(-1,1)$.
I know that $\int_{-1}^1\cos(nt)dt\to 0$, as $n\to\infty$, and I tried to extract this term as a multiplicand (applying Holder's inequality) but with no success.
Or maybe I should take into account that $L_2(-1,1)$ is a Hilbert space and somehow  use the general form of continuous linear functionals there? 

Two ways (or more) to solve the problem: 
First one: with analysis, using the fact that the set of continuously differentiable functions is dense in $L_2(-1,1)$. Then take $x\in L_2(-1,1)$, and $\{y_k\}\subset C^1(-1,1)$ which converges to $x$ in $L_2(-1,1)$. Then 
\begin{align*}
|f_n(x)|&\leq |f_n(x-y_k)|+|f_n(y_k)| \\
&\leq \sqrt 2||x-y_k||_{L^2}+ \left|\frac{y_k(1)\sin(n)-y_k(-1)\sin(-n)}n\right|+\left|\int_{-1}^1y_k'(t) \frac{\sin(nt)}ndt\right|\\
&\leq \sqrt 2||x-y_k||_{L^2}+\frac{|y_k(1)|+|y_k(-1)|}n+\frac 2n\sup_{-1\leq t\leq 1}|y_k'(t)|
\end{align*}
 and for all $k$: $\limsup_n|f_n(x)|\leq \sqrt 2||x-y_k||_{L^2}$ so you can conclude.
Second one: using only measure theory facts: 


*

*Prove that $$\mathcal B:=\left\{A\subset (-1,1),\lim_{n\to \infty}\int_A\cos(nt)dt=0\right\}$$
is a $\sigma$-algebra. 

*Conclude that for all Borel-measurable $B\subset (-1,1)$ we have $\lim_{n\to \infty}\int_B\cos(nt)dt=0$.

*Show that for all simple function $s$ we have $\lim_{n\to \infty}\int_{(-1,1)}s(t)\cos(nt)dt=0$.

*Conclude.

