# On bounded finite-rank operators (Hilbert spaces)

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.

• The range of a linear map is a vector space. You can take dimensions of vector spaces. What is the problem here? Jan 19 at 15:25
• I'm just confused with the definition of "dimension". I used to look at the number of components of the vector in that space, but here I have functions so... what exactly is a component here? Jan 19 at 15:35
• That's a terrible way. For example, $\{(\lambda,0): \lambda\in \mathbb{R}\}$ is one-dimensional but vectors here have two components. Am I misunderstanding what you say? Jan 19 at 15:37
• No, you're not. So for example $e^{ikx}$ is a basis for $L([0,1])$. What matters is the number of "vectors" $e^{ikx}$. They are actually the "components" with which I construct every other vector (square-integrable function on $[0,1]$). Jan 19 at 15:44
• What is your definition of basis? Because that is not a Hamel basis (but rather an orthonormal basis). Jan 19 at 15:45

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 $$\operatorname{dim} \operatorname{ran} T \leq n$$.