Let $X$ be a normed space, and let $T\in B(X)$ be a bounded operator with $\mathrm{dim}R(T)=n\lt\infty$.I'm trying to prove that there exists linealy independent vectors $y_1,\cdots,y_n$ in $X$ and linearly independent functions $f_1,\cdots,f_n$ in the dual space $X^*$ of bounded linear functionals such that $T=y_1\otimes f_1+\cdots +y_n\otimes f_n$.(The tensor product $y\otimes f$ is defined by $(y\otimes f)(x)=f(x)y$.)
Here is my attempt:
Let $y_1,\cdots,y_n$ be a basis for $R(T)$. Then for each $x\in X$, there is a unique list of scalars $\alpha^{(x)}_1,\cdots,\alpha^{(x)}_n$ such that $Tx=\sum_{i=1}^n{\alpha^{(x)}_i y_i}$. And for each $1\leq i\leq n$, let $f_i: X\rightarrow\mathbb{F}$ be the function defined by $f_i(x)=\alpha^{(x)}_i$. Then we obtain $Tx=\sum_{i=1}^n{f_i(x) y_i}$ for all $x\in X$.
I have showed that each $f_i$ is indeed a linear functional. However, I have no idea whether they're bounded or not. I was given a way to complete this proof by considering two equivalent norms $\parallel Tx\parallel$ and $\sum_{i=1}^n{|f_i(x)|}$ on $R(T)$. However, I do not think they are necessarily norms on $R(T)$! For instance, if $x\in R(T)$ and $\parallel Tx\parallel=0$, then how to show that $x=0$ ?
Could anyone help to finish this proof? Thanks in advance.