Approximation Property: Decomposition

Question

This thread originated from MSE: Approximation Property: Decomposition

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.)

Answer 1

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.