Is every finite rank operator a linear combination of rank one projections? Let $X$ be a Banach space and let $F(X)\subseteq B(X)$ be the subset of finite rank operators. Let $u\in F(X)$, must $u$ be a linear combination of rank one projections? 
I know this result is true for Hilbert spaces, but I suspect it fails in general even though I don't have a counterexample.
 A: I believe yes.
Let $X$ be Banach and $A: X\to X$ finite rank. We will define sub-spaces $\mathcal{I, K, C}$ of $X$. $\mathcal I$ is the image of $A$, it is finite dimensional. $\mathcal K$ is the kernel of $A$ and it is closed, we choose a complement $\mathcal C$ of $\mathcal K$, this complement is finite dimensional.
Now $A$ mainly lives on the space $\mathcal C + \mathcal I$, which is a finite dimensional sub-space of $X$. Denote the restriction $A\lvert_{\mathcal C+\mathcal I}: \mathcal C+\mathcal I\to\mathcal C+\mathcal I$  by $B$. As $B$ is a map between finite-dimensional vector spaces we find that $B$ is a linear combination of projections $p_i$ living in $\mathcal C+\mathcal I$, write $B=\sum_i \lambda_i p_i$.
The goal of what comes is to find a continous projection $\overline P: X\to X$ so that $\overline P$ has image $\mathcal{C+I}$ and so that if $x\in\mathcal K$ then $\overline{P}(x)\in\mathcal I\cap\mathcal K$. If we have such a projection, then denoting the inclusion $\mathcal C+\mathcal I \to X$ by $\iota$ we define
$$\overline A :=\sum_i \lambda_i\, (\iota\circ p_i\circ \overline P).$$
Basically $\overline A(x)=B(\overline P(x))$. It is elementary to check that $\iota\circ p_i\circ\overline P$ is a projection. Also if $x\in\mathcal C$ then $\overline P(x)=x$ and then $\overline A(x)=B(\overline P(x))=B(x)=A(x)$. If $x\in\mathcal K$ then $\overline P(x)\in\mathcal I\cap\mathcal K$ and $\overline A(x) = B(\overline P(x))=0=A(x)$. This then gives the result.
We proceed in two steps. First choose a basis $e_1,.., e_n$ of $\mathcal C$ and a basis $\{b_\xi\}_{\xi\in J}$ of $\mathcal K$. Define $P:X\to X$ by $P(\sum_i x_i e_i + \sum_\xi y_i b_\xi)=\sum_i x_ie_i$. This is a projection on $X$ with kernel $\mathcal K$ and image $\mathcal C$, both of which are closed. By comparing with this question we note that $P$ is continuous.
We will now introduce another big projection. Let $d_1,...,d_m$ be a basis of $\mathcal I\cap \mathcal K$. Extend the functionals $d_i^*:\mathcal I\cap \mathcal K\to \Bbb C$ defined by $d_i^*(d_j)=\delta_{ij}$ to all of $\mathcal K$ with Hahn-Banach. Then define $P'(x) = \sum_i d_i\cdot d_i^*(x)$. It is immediate that $(P')^2=P'$ and $P'$ has image $\mathcal I \cap \mathcal K$, giving us a continuous projection $P':\mathcal K\to \mathcal K$ with $P'(\mathcal K)=\mathcal I\cap\mathcal K$. We further define $P'':X\to X$ by letting $P''(\mathcal C)=0$, and $P''\lvert_{\mathcal K} = P'$. The image of $P''$ is $\mathcal I\cap \mathcal K$, which is closed, and the kernel is $\ker(P')\oplus\mathcal C$, which is also closed. Thus $P''$ is continuous projection.
Finally define $\overline{P}=P+P''$. The image of $\overline{P}$ is $\mathcal C\oplus \mathcal I\cap\mathcal K= \mathcal C + \mathcal I$, and if $x\in\mathcal K$ then $\overline P(x) = P''(x) \in\mathcal I \cap\mathcal K$ as desired.
