I'm having trouble understanding how these two representations can coexist. Citing Wikipedia:
In functional analysis, a branch of mathematics, a finite-rank operator is a bounded linear operator between Banach spaces whose range is finite-dimensional.
Finite-rank operators are matrices (of finite size) transplanted to the infinite-dimensional setting.
$(\dots)$ an operator $T$ of finite rank $n$ takes the form $$Th=\sum _{{i=1}}^{n}\alpha _{i}\langle h,v_{i}\rangle u_{i}\quad {\mbox{for all}}\quad h\in H$$
where $\{u_i\}$ and $\{v_i\}$ are orthonormal bases. $(\dots)$ This can be said to be a canonical form of finite-rank operators.
For $H=\ell^2$ it makes sense but… what if we have, say, $H=L^2([0,1])$ or $L^2(\mathbb{R})$? These are function spaces. How can the range $\operatorname{ran} T$ be finite-dimensional? At the end of the day, it sends functions to functions, and they are infinite-dimensional objects.
Best Answer
Dimension is defined for a vector space $V$ as the cardinality of a basis of $V$. In this context it makes no sense to say, a function is an infinite-dimensional object. What is infinite-dimensional is the space of functions $L^2((0,1))$ for example.
In an Hilbert space $H$ we could take a finite-dimensional subspace $U \subset H$ and the projection operator $P$ projecting $H$ onto the subspace $U$. Then $\operatorname{ran} T = U$, so $T$ is a finite rank operator.
If $X,Y$ are Banach spaces we could also take any linear functionals $x_i' \in X'$ and vectors $y_i \in Y$ and define an operator $T: X \to Y$ by $Tx := \sum_{i=1}^n x_i'(x)y_i$. $T$ is a finite rank operator with $\operatorname{ran} T \subset \operatorname{span}(y_i)$, so $\operatorname{dim} \operatorname{ran} T \leq n$.