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?
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?