Let $X$ be a Banach space.Then for every nonzero $x\in X$, and nonzero $f\in X^*$, I'm told that the tensor product $x\otimes f: X\rightarrow X$ defined by $(x\otimes f)y=f(y)\,x$ is a rank-one operator and every rank-one operator can be written in this way.
I have showed that such nonzero linear functional $f$ must be surjective and thus the range of $x\otimes f$ is $\mathrm {span}(x)$. But why is every rank one operator of the form $x\otimes f$? Could anyone please tell me a solution or reference?
Let $\ell$ be a bounded rank-one linear functional. Then by definition, its range is of dimension $1$, so there exists $x\neq 0$ such that it is $\mathrm{span}(x)$. Hence for any $y\in X$, there exists a scalar (say $f(y)$) such that $\ell(y) = f(y)\,x$. Since $\ell$ is linear and bounded, so is $f$, which proves $f\in X^*$.