# canonical form of finite rank operators

Let $X,Y$ be banach spaces and let $T:X\rightarrow Y$ be an linear continuous operator with finite dimensional image $Im(T)\subset Y$. Now I want to prove that there exists continuous linear functionals $x_i':X\rightarrow \mathbb{R}, x_i'\in X'$ and vectors $y_i\in Y$ such that $T(x)= \sum\limits_{i=1}^n x_i'(x)\cdot y_i$ for all $x\in X$.

I need help in proving this. Suppose we have an basis $\lbrace y_1,...,y_n\rbrace$ for the image $Im(T)$. This should be possible, since the image has finite dimension. Now we know that the image $Tx$ of any vector $x\in X$ must be some linear combination of $y_1,...,y_n$. Therefore we can write:

$Tx=\sum\limits_{i=1}^n f_i(x)\cdot y_i$ for some real valued functions $f_i:X\rightarrow \mathbb{R}$. My idea was to put $x_i':=f_i$. I could prove that the functions $x_i'=f_i$ are linear. But I have trouble to prove they are indeed continuous. Do you have an idea? Or is my approach not good enough?

Best regards

You can define linear functionals $g_j:Im(T) \to \mathbb C$ by $$g_j(\sum_{i=1}^n a_i y_i)=a_j.$$ Another way to phrase this is that $g_j$ is defined by setting $$g_j(y_i)=\delta_{ij}$$ and then extending linearly. Using the fact that all norms on finite-dimensional spaces are equivalent, this can easily be shown to be continuous.
Then $g_j \circ T$ is a continuous linear functional on $X$ (as it is a composition of linear maps). But $$(g_j \circ T)(x) = \sum_{i=1}^n f_i(x) g_j(y_i) =f_j(x).$$ So your linear map $f_j$ is equal to $g_j$ and so must be continuous.