Range of a degenerate integral operator is closed

Question

I am reading around about integral operators and I came across an interesting example that I could not figure out. This example comes from here. It is Example 5.27, I have typed it verbatim for your utility (I hope this is legal, I don't mean to infringe on anyone's work!)

Example 5.27 An integral operator $K: C([0,1])\to C([0,1])$ $$Kf(x) = \int_0^1k(x,y)f(y)dy$$ is said to be degenerate if $k(x,y)$ is a finite sum of separated terms of the form $$ k(x,y) = \sum_{i=1}^n\varphi_i(x)\psi_i(y),$$ where $\varphi_i,\psi_i:[0,1]\to \mathbb{R}$ are continuous functions. We may assume without loss of generality that $\{ \varphi_1,\ldots,\varphi_n \}$ and $\{ \psi_1,\ldots, \psi_n\}$ are linearly independent. The range of $K$ is the finite-dimensional subspace spanned by $\{ \varphi_1,\varphi_2,\ldots,\varphi_n \}$, and the kernel of $K$ is the subspace of functions $f \in C([0,1])$ such that $$\int_0^1f(y)\psi_i(y)dy = 0 ~~~~\text{ for } i=1,\ldots,n. $$ Both the range and kernel are closed linear subspaces of $C([0,1])$.

I see that the kernel and range are subspaces, and I see that the kernel is closed (Theorem 5.25 from the same text), but I cannot figure out how they came to the conclusion that the range is closed.

If someone could explain to me how the authors came to this conclusion I would be most grateful.

Answer 1

Since $Kf=\sum\limits_{i=1}^n\left(\int\limits_0^1\psi_i(y)f(y)dy\right)\varphi_i$, the range of $K$ is contained in the span of $\{\varphi_1,\ldots,\varphi_n\}$ as claimed.

Every finite dimensional subspace of a normed space is closed. Every finite dimensional normed space is complete. All norms are equivalent on a finite dimensional complex vector space. (See this related question.) Proofs of all of this are found in Section 5.4 of the file you linked to, which comes shortly after your example.