I want to show that if a normed linear space has Radon-Riesz property, then it has the semi Radon-Riesz property.
We know that a normed linear space is said to have Radon-Riesz property if for every sequence $(x_n)$ in $X$ such that $x_n\xrightarrow{\text{weakly}}x$ in $X$ and $\|x_n\|\to \|x\|$, then $x_n\to x$.
We also know that a normed linear space is said to have semi Radon-Riesz property if for a sequence $(x_n)$ in $S_X, x\in S_X$ and $(x_n^*)$ in $S_{X^*}$, $x_n\xrightarrow{\text{weakly}}x$ and $x_n^*(x_n)=1$ for all $n\in \mathbb{N}$, then $x_n^*(x)\to 1$.
It is also easy to show that a normed linear space has Radon-Riesz property if and only if for a sequence $(x_n)$ in $S_X$ and $x\in S_X$, $x_n\xrightarrow{\text{weakly}}x\implies x_n\to x$.
Let $X$ have Radon-Riesz property and let $(x_n)$ be a sequence in $S_X$, $x\in S_X$ and $(x_{n}^*)$ a sequence in $S_{X^*}$ such that $x_n\xrightarrow{\text{weakly}}x$ and $x_n^*(x_n)=1$ for all $n\in \mathbb{N}$. Now how to show that $x_n^*(x)\to 1$? I am stuck here. Any help will be appreciated.