Weakly sequentially continuous maps

Question

I have come across this map $F:H^2(0,1)\to L^2(0,1)$, where $F(w)=w^3$. I want to show that the map is weakly sequentially continuous. I started with

Pick a sequence $w_n$ weakly convergent to some $w_0$ in $H^2(0,1)$. Since $H^2(0,1)\subset \subset L^6(0,1)$, the sequence $w_n^3$ strongly converges to an element $v_0^3$ in $L^2(0,1)$.

Now how can I show $v_0=w_0$?

Answer 1

You have $w_n \to w$ in $L^6$. Then $$ w_n^3-w^3= (w+w_n-w)^3-w^3 = 3w(w-w_n)^2 + 3(w-w_n)w^2 $$ should help to conclude $w_n^3 \to w$ in $L^2$. Alternatively, you could try to apply dominated convergence.

You could also use the compact embedding $H^2(0,1)$ into $C([0,1])$, which seems to be much easier.