Let $V$ be a Banach space and we say that $V$ has the $C$-BAP if there exists a net of bounded finite rank operators $T_\alpha$ in $B(V,V)$ and a constant $C$ such that $\|T_\alpha\| \leq C$ for each $\alpha$ and $\|T_\alpha(y)-y\| \rightarrow 0$ for each $y \in W$. I am trying to prove the following:
Show that if $V$ is separable, the BAP can be described by a sequence of bounded finite rank operators $\{T_n\}$, i.e. there exists a sequence of finite rank operators $\{T_n\}$ on $V$ and a constant $C$ such that $\|T_n\| \leq C$ and $\|T_n(y)-y\| \rightarrow 0$ for every $y \in V$.
My attempt: I was trying to solve this problem along the lines of this property that if $V$ is separable, then the closed unit ball of $V^\ast$ is weak$^\ast$ metrizable. But here we are dealing with operators whose image is also in $V$ unlike the scalar field $\mathbb{F}$ for linear functionals.
Can anyone help me?
Let $\{v_n\}$ be a dense countable subset of $V$. Then we can find: $$ \alpha_1,\ \ \ \text{ such that } \|T_{\alpha_1}(v_1)-v_1\|<1. $$ $$ \alpha_2,\ \ \ \text{ such that } \|T_{\alpha_2}(v_j)-v_j\|<\frac12,\ \ j=1,2. $$ $$ \alpha_3,\ \ \ \text{ such that } \|T_{\alpha_3}(v_j)-v_j\|<\frac13,\ \ j=1,2,3. $$ $$ \vdots $$ $$ \alpha_n,\ \ \ \text{ such that } \|T_{\alpha_n}(v_j)-v_j\|<\frac1n,\ \ j=1,\ldots,n. $$ Etc.
Write $T_n=T_{\alpha_n}$. By construction, $\|T_n(v_j)-v_j\|\xrightarrow[n\to\infty]{}0$ for all $j$ .
Now, for arbitrary $v\in V$ and $\varepsilon>0$, there exists $j$ with $\|v_j-v\|<\varepsilon/(2C+2)$, and $n_0$ such that $\|T_n(v_j)-v_j\|<\varepsilon/2$ for all $n>n_0$. Then \begin{align} \|T_n(v)-v\|&\leq\|T_n(v)-T_n(v_j)\|+\|T_n(v_j)-v_j\|+\|v_j-v\|\\ \ \\ &\leq(C+1)\|v_j-v\|+\|T_n(v_j)-v_j\|\\ \ \\ &\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon. \end{align}