*This thread originated from MSE: [Approximation Property: Decomposition](https://math.stackexchange.com/q/1124860/79762)*


Given a Banach space $E$.


Consider a finite rank operator $F\in\mathcal{F}(X,E)$.


Introduce a basis on the finite dimensional range:
$$\dim\mathcal{R}F<\infty:\quad y\_1,\ldots, y\_N$$
Hahn-Banach lifts that dual basis up:
$$ y\_n\in Y':\quad\langle y\_n',y\_m\rangle=\delta\_{mn}$$
So the finite rank operator decomposes into:
$$Fx=\sum\_{n=1}^N\langle y\_n,Fx\rangle y\_n=\sum\_{n=1}^N\langle F'y\_n',x\rangle y\_n=\left(\sum\_{n=1}^Ny\_n\otimes x\_n'\right)x\quad(x\_n':=F'y\_n')$$
*(As expected it has a representation as a sum.)*


Given a Hilbert space $\mathcal{H}$.


Consider a compact operator $C\in\mathcal{C}(X,\mathcal{H})$.


Introduce a basis on the separable Hilbert space:
$$\dim\overline{\langle C(B)\rangle}\leq\mathfrak{n}:\quad\varepsilon\_1,\ldots$$
One obtains a series of increasing projections:
$$P\_N\varphi:=\sum\_{n=1}^N\langle\varepsilon\_n,\varphi\rangle\varepsilon\_n:\quad\|P\_NC-C\|=\|P\_N-1\|\_{C(B)}\stackrel{N\to\infty}{\to}0$$
So the compact operator decomposes into:
$$C\varphi=\sum\_{n=1}^\infty\langle\varepsilon\_n,C\varphi\rangle\varepsilon\_n=\sum\_{n=1}^\infty\langle C^\*\varepsilon\_n,\varphi\rangle\varepsilon\_n=\left(\sum\_{n=1}^\infty\varepsilon\_n\otimes\delta\_n^\*\right)\varphi\quad(\delta\_n^\*:=C^\*\varepsilon\_n^\*)$$
*(As hoped it has a representation as a series.)*


Back to a Banach space $E$.



> 
> What about almost finite rank operators between Banach spaces?
>  $$C\in\overline{\mathcal{F}(X,E)}\subseteq\mathcal{C}(X,E):\quad Cx=\left(\sum\_{n=1}^\infty y\_n\otimes x\_n'\right)x$$
>  *(Clearly, for compact but not almost finite rank ones this can't hold.)*
> 
> 
> 



Approximation Property: Decomposition
You can always get such a representation. First, given $C$ in the closer of the finite rank operators, you can write it as an infinite sum $\sum T\_n$ of finite rank operators (even with $\|T\_n\| < 2^{-n}$ for $n>1$). Let $E\_n$ be the range of $T\_n$, with dimension $m(n)$, say. By Pelczynski's argument, you can write the identity on $E\_n$ as the sum of $m(n)^2$ rank one operators so that all the partial sums have norm at most $2$. What to do next should be clear. 


A. Pelczynski, Any separable Banach space with the bounded approximation property is a complemented subspace of a Banach space with a basis. Studia Math. 40 (1971), 239–243. 


