I am currently doing an exercise on compact operators.
Problem
Let $K: X \to Y$ be a compact linear operator. Suppose $(x_{n})_{n \in \mathbb{N}}$ is a sequence in $X$ with the property that there exists $x \in X$ such that $f(x_n) \longrightarrow f(x)$ for each bounded linear functional $f$ in $X^{*}$.
a) Show that $Kx_n \longrightarrow Kx$
My solution
$$\|Kx_n-Kx\| \leq \|K\|\times\|x_n-x\|$$
In the above equation, I have used the fact that $K$ is compact and compact operators are bounded. Now a corollary of the Hahn-Banach theorem is the following:
Corollary
For any $x \in X$, where $X$ is a normed linear space. There exists a linear functional $f \in X^{*}$ such that $\|f\| = 1$ and $f(x) = \|x\|$
Now since $X$ is a vector space, $x_n-x \in X$, thus by the above corollary there exists a bounded linear functional $f \in X^{*}$ such that $f(x_n - x)= \|x_n-x\|$.
We then have $$0 \leq \|Kx_n-Kx\| \leq \|K\|\times\|x_n-x\| = \|K\|(f(x_n)-f(x)) \leq \|K\| |f(x_n)-f(x)| $$
Thus when we take the limit, $n \longrightarrow \infty$, by our hypothesis we have $|f(x_n)-f(x)|\longrightarrow 0$. So we see by virtue of the squeeze theorem that $\|Kx_n-Kx\| \longrightarrow 0$. Hence $Kx_n \longrightarrow Kx$.
Now I have a doubt. I don't see the need for $K$ to be compact, it suffices to have that $K$ is bounded. Furthermore, we know that not every bounded linear operator is compact. Its possible I made an error in my calculations. If you may be generous enough could you please point it for me?
