Let $0<\sigma< \frac{2}{N-2}$ with $N \geq 3$. I know that $H^{1}_{\operatorname{rad}}(R^n)$ (radial functions of $H^{1}(R^n)$ ) is compactly embedded in $L^{2 \sigma +2}(R^N)$.
Let $(\psi_v )$ a bounded sequence in $H^{1}(R^n)$ of radial function.
The article that i am studying says :
Since the sequence $(\psi_v )$ is bounded in $H^{1}(R^n)$ , some subsequence has a weak $H^1$ limit $\psi^{*}$ (because $H^{1}(R^n)$ is reflexive). Since $(\psi_v )$ are radial and uniformly bounded in $H^{1}(R^n)$, it follow by the compact immersion (the immersion that i said before) that we can take $(\psi_v )$ strongly convergent to $\psi^{*}$ in $L^{2 \sigma +2}(R^N)$.
I am not understanding the last affirmation... someone can give me a help ?
thanks in advance
Fact. If $X$ and $Y$ are Banach spaces, $T : X\to Y$ is a compact map, and $x_n\to x$ weakly in $X$, then $Tx_n\to Tx$ strongly in $Y$.
Weakly means, $x^*(x_n)\to x^*(x)$, for every $x^*\in X^*$.
How to prove this Fact:
It suffices to show that every subsequence of $Tz_n=T(x_n-x)$ possesses a sub-subsequence which converges to $0$.
Let $Ty_n$ be a subsequence of $T(x_n-x)$. Clearly $y_n$ is bounded, as $x_n$ is also bounded, since it converges, and as $T$ is compact, then $Ty_n$ possesses a converging subsequence $\{Ty_{n_k}\}_{k\in\mathbb N}$, with $Ty_{n_k}\to w$. We shall show $w=0$.
We now define $T^*: Y^*\to X^*$ the adjoint of $T$, as $$ (T^*y^*)(x)=y^*(Tx). $$ Clearly $T^*$ is also bounded, as $$ |T^*(y^*)(x)|=|y^*(Tx)|\le |y^*|\, |Tx|\le |y^*|\, |T|\, |x|, $$ and as $Ty_{n_k}\to w$, then $$(T^*y^*)(y_{n_k})=y^*(Ty_{n_k})\to y^*(w),$$ for every $y^*\in Y^*$.
But $x^*(y_{n_k})\to 0$ and thus $y^*(w)\to 0$, for all $y^*\in Y^*$, which means that $w=0$.