Let $A: C[0,1] \rightarrow C[0,1], Af(x) = f(x)/x$ with domain $D(A) = ${$f \in C[0,1]: \exists \lim_{x \rightarrow 0^+} f(x)/x$}.
Prove that $A$ is closed.
I know that A linear operator $T:X \rightarrow Y$ is closed provided whenever {$x_n$} $\rightarrow x_0$ and $T(x_n) \rightarrow y_0$ then $T(x_0) = y_0$, but I'm not really sure how to proceed. I see a theorem that says a linear operator between Banach spaces is continuous iff its closed, but I'm not really sure how to apply that here or if I even should
Suppose $f_n \to f$, $A f_n \to g$, we want to show that $Af = g$.
The idea is note that if we let $\gamma(x) = x g(x)$, then $A \gamma = g$ and all we need to do is to show that $f_n \to \gamma$.
Since $Af_n \to g$, for any $\epsilon>0$ then there is some $N$ such that $|{f_n(x)\over x} - {g(x)}| \le \epsilon$ for all $x \neq 0$ and $n \ge N$. Hence $|f_n(x)-xg(x)| \le \epsilon|x| \le \epsilon$. In particular, $\|f_n-\gamma\| \to 0$, where $\gamma(x) = x g(x)$.
Hence $f= \gamma$ and so $A$ is closed.