I'm with trouble to solve (b).
(a) Let $W$ be a Banach space. Assume that there exists a sequence $\{P_n\}\subset\mathcal{B}(W,W)$ of finite rank operators such that $P_n(y)\rightarrow y$, for all $y\in W$. Show that, if $V$ is a Banach space and $T\in\mathcal{B}(V,W)$ is compact, then $T$ is the limit of finite rank operators.
(b) Deduce that, if $V$ is any Banach space and $T\in\mathcal{B}(V,\ell^p)$ ($1\leqslant p<\infty$) is compact, then $T$ is the limit of finite rank operators.
In the (a) part, I just define $T_n:=P_n\circ T$, for each $n\in\Bbb{N}$, and get the desired, because:
- $R(T_n)\subset R(P_n)$ implies that $\dim R(T_n)\leqslant\dim R(P_n)<\infty$, since $P_n$ has finite rank, for all $n$.
- Since $P_n(y)\rightarrow y$, for all $y\in W$, given $\varepsilon>0$, there exists $N_0\in\Bbb{N}$ such that $$\|P_n(y)-y\|<\varepsilon,\quad\forall n\geqslant N_0.$$ Now, for any $v\in B_V[0;1]$, we have that for all $n\geqslant N_0$ $$\|(T_n-T)(v)\|=\|T_n(v)-T(v)\|=\|P_n(\underbrace{T(v)}_{\in W})-T(v)\|<\varepsilon,$$ which implies that $$\sup_{v\in B_V[0;1]}\|(T_n-T)(v)\|=\|T_n-T\|<\varepsilon.$$
Obviously, to solve (b), I just need to find the sequence $\{P_n\}$ and apply (a), but I can't see how.
Best Answer
Just take $P_n$ to be the projection into the first $n$ terms, i.e. $P_n:(a_0, a_1, \dotsc)\mapsto (a_0, a_1, \dotsc, a_{n - 1}, 0, 0, \dotsc)$.
Then $P_n$ is a finite rank operator, because its image is $n$-dimensional. And it's routine to check that $P_n(y)\rightarrow y$ for any $y\in \ell^p$.
However, I think your proof for (a) is not quite correct. The problem is that your $N_0$ depends on the choice of $y$, while later you used it for an arbitrary vector (namely $T(v)$) in $W$. The "fix" is to use the compactness assumption of $T$, so that a universal $N_0$ can be achieved.