Suppose I am given that $X$ is a normed space and $(x_{n})_{n=1}^{\infty}$ is a sequence in $X$ such that $(f(x_{n}))_{n=1}^{\infty}$ is convergent in $K$ for any $f\in X^{*}$ where $K$ is either real or complex and $X^{*}$ is the dual space of $X$. Show that there exists $F\in X^{**}$ so that $F(f)=\underset{n}{\lim}f(x_{n})$ for all $f\in X^{*}$
Do I prove the existence of $F$? Isn't it defined in the question? How do I use the convergence part?
I am terrible at this so some help will be greatly appreciated.
Just define the linear map $F$ as $F(f)=\lim_{n}f(x_{n})$ and try to show that $F$ is bounded. Indeed, let $F_{n}(f)=f(x_{n})$ for $f\in X^{\ast}$, $n=1,2,...$, then $\lim_{n}F_{n}(f)$ exists by assumption, so the result follows by Uniform Boundedness Principle since $\sup_{n}F_{n}(f)<\infty$ for each $f$, and hence $\sup_{n}\|F_{n}\|<\infty$, so $|F(f)|=\lim_{n}|f(x_{n})|=\lim_{n}|F_{n}(f)|\leq\sup_{n}\|F_{n}\|\|f\|$, so $\|F\|\leq\sup_{n}\|F_{n}\|<\infty$.