Finite Rank Operator in Normed Space, not necessarily Hilbert neither Banach

Question

Suppose that $E$ and $F$ are normed spaces and $T:E \rightarrow F$ is a bounded linear operator.

I NEED TO SHOW WHAT FOLLOWS:

If there are $n\in \mathbb{N}, f_{1}, ..., f_{n}\in E^{\ast}$ (dual of $E$) and $y_{1}, ..., y_{n}$ such that $T(x)=f_{1}(x)y_{1}+ ... + f_{n}(x)y_{n}$, for all $x\in E$, then $T$ has finite rank (its imagem has finite dimension).

Please, if anyone can help me with that. I couldn't find anything beside results for Hilbert spaces...

Here I can't use arguments with adjoint operators, compact or such things.

I see that $y_{1}, ... , y_{n}$ span the Image of $T$, but I dont know how to show that they are in its image. Cuz they dont need to be, at first.

Answer 1

You have by hypothesis that $T(x)$ can be written as a linear combination of $y_1,\dots,y_n\in F$ for every $x\in E$.

Thus, $T(E)\subseteq{\rm span}\,\{y_1,\dots,y_n\}$ and therefore $T(E)$ is a subspace of ${\rm span}\,\{y_1,\dots,y_n\}$, which is finite dimensional.

You're right when you say that $y_j$ need not to be in $T(E)$, but that doesn't matter.