How to study the strong and weak convergence of this sequence of functions? Discuss the strong and weak convergence of the sequence
$$u_n(x)=\sin(x)+\frac{1}{n}\sin^2(nx)$$
in the Sobolev space $W^{1,2}(0,1)$.
I know that a function $u(x)$ belong to $W^{1,p}(a,b)$ iff the $L^p$ norms of $u(x)$ and $u'(x)$ are finite. In particular we can express the norm in a Sobolev space in this way 
$$||{u}||_{W^{1,p}}=||u||_{L^p}+||u'||_{L^p}$$
where
$$||u||_{L^p}^p=\int_\Omega|u|^pd\omega$$
Since the following integrals are hard to compute i think there is another way to discuss this problem.
$$||u_n||_{L^2}^2=\int_0^1|\sin(x)+\frac{1}{n}\sin^2(nx)|^2dx$$
$$||u'_n||_{L^2}^2=\int_0^1|\cos(x)+2\cos(nx)\sin(nx)|^2dx$$
Maybe could be helpfull knowing that
$$\lim_{n\rightarrow\infty}u_n(x)=\sin(x)$$
Could someone help me?
 A: The $L^2$ norm of $u_n(x)$ is not so hard to deal with. Remember, we just have to show that it is finite:
\begin{align}
\int_0 ^1\left|\sin(x)+\frac{1}{n}\sin^2(nx)\right|^2\;dx &\leq \int_0^1\left(|\sin(x)| + \frac{1}{n}\sin^2(nx)\right)^2\;dx\\
&=\int_0^1\sin^2(x)+\frac{2}{n}|\sin^3(x)|+\frac{1}{n^2}\sin^4(x)\;dx \\
&\leq \int_0^1 1+\frac{2}{n}+\frac{1}{n^2}\;dx \\
&=1+\frac{2}{n}+\frac{1}{n^2} \\
&< \infty
\end{align}
So $u_n(x)\in L^2((0,1))$. You can do similarly for $u_n'(x)$. Now we look at weak convergence in $L^2$. You made the guess $u_n\rightharpoonup \sin(x)$. We have to show that:
$$\lim_{n\rightarrow\infty}\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx=0,\qquad\forall\phi\in L^2$$
We can show this by finding an upper bound on the magnitude of this integral:
\begin{align}
\left|\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx\right| &=\left|\int_0^1 \frac{1}{n}\sin^2(nx)\phi(x)\;dx\right|\\
&\leq\frac{1}{n}\int_0^1\sin^2(x)\left|\phi(x)\right|\;dx\\
&\leq\frac{1}{n}\int_0^1\left|\phi(x)\right|\;dx \\
&= \frac{M}{n}
\end{align}
where $M = \int_0^1|\phi(x)|\;dx <\infty$ because $\phi\in L^2$. Taking the limit, we obtain:
$$\lim_{n\rightarrow\infty}\left|\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx\right|= 0$$
For weak convergence of $u_n'$, we make use of the fact that step functions are dense in $L^2$, so that for any $\phi(x)\in L^2$, there exists a sequence of step functions $\phi_\nu(x)$ that converges strongly to $\phi(x)$ in $L^2$. This allows us to just consider the integral of $[u_n'-u']\phi_\nu$ on each interval $[c_i,c_{i+1})$ on which $\phi_\nu$ is constant:
\begin{align}
\int_{c_i}^{c_i+1}[u_n'(x) - u'(x)]\phi(x)\;dx &= \int_{c_i}^{c_i+1} \sin(2nx)\phi(x)\;dx\\
&=\int_{c_i}^{c_i+1} \sin(2nx)\phi_\nu(x)\;dx + \int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx \\
&=\phi_\nu(c_i)\int_{c_i}^{c_i+1}\sin(2nx)\;dx+\int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx \\
&= \phi_\nu(c_i)\frac{\cos(2nc_{i}) - \cos(2nc_{i+1})}{2n} +\int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx 
\end{align}
Strong convergence of $\phi_\nu$ to $\phi$ means we can make the second integral arbitrarily close to zero. Taking the limit as $n\rightarrow \infty$:
$$\lim_{n\rightarrow \infty}\int_{c_i}^{c_i+1}[u_n'(x) - u'(x)]\phi(x)\;dx =0$$
So $u_n\rightharpoonup \sin(x)$ in $W^{1,2}$. Can you deal with strong convergence?
