strong and weak* limits

Question

I have the sequence $\Phi_n(x)=\sin(nx) \ \forall x\in(0,\pi),n\in \mathbb{N}$ and I know that it weakly* converges to $0$ in $L^{\infty}((0,\pi))$. Now I have to show that it doesn't converge strongly in $L^{1}((0,\pi))$. Can I say that if it should converge in $L^{1}((0,\pi))$ it must converges to the same limit of the weak* convergence? or since the space is different also the limits can be different?

Answer 1

$\int_{n\pi}^{(n+1)\pi} |\sin x |dx =\int_{0}^{\pi} |\sin (y+n\pi) |dy=\int_0^{\pi} |\sin y|dy$. Call this fixed positive number $A$. By Cesaro's Theorem the averages of $(\int_{n\pi}^{(n+1)\pi} |\sin x |dx)$ also tend to $A$. This means $\frac 1 n \int_0^{n\pi} |\sin x |dx \to A$. By a change of variable this gives $\int_0^{\pi} |\Phi_n (x)|dx \to A$. Suppose $\Phi_n \to g$ in $L^{1}$. Then there is a subsequence $(\Phi_{n_k})$ which converges to $g$ a.e. By DCT $\int \Phi_{n_k} f \to \int fg$ for all $f \in L^{1}$. Hence, $\Phi_{n_k} \to g$ in $weak^{*}$ topology. By uniqueness of $weak^{*}$ limits we get $g=0$ a.e. But then $\Phi_n \to 0$ in $L^{1}$ which is contradiction to the fact that $\int_0^{\pi} |\Phi_n (x)|dx \to A>0$.