# Showing that the Hilbert-Schmidt integral operator is compact.

I am wondering if my proof sketch works for showing that the Hilbert-Schmidt integral operator is compact. I also have a few parts that I am having some difficulty with, and would appreciate help and feedback on them.

We define a Hilbert-Schmidt integral operator $T$ over domain $X$ with $k \in L^2 (X \times X)$:

$$T[f](t) = \int_{X} k(t, s) f(s) \ d\mu(s)$$

The sketch goes as follows:

1. Since integrable simple functions are dense in $L^2$, let us approximate $k$ with some sequence of simple functions $\{ k_n\}$ where:

$$\lim_{n \to \infty} \int_{X \times X} | k - k_n| ^2 \ d(\mu \times \mu)= 0$$

2. Note that each $k_n$ has form:

$$k_n (x, y) = \sum_{i = 1}^{N(n)} k_{n, i} \mu(E_{n, i}) \chi_{E_{n, i}} (x, y)$$

With the partition size $N(n) < \infty$ for all $n$, and the measure $\mu$ being Lebesgue. Let us define $T_n$ as:

$$T_n [f](t) = \int_{X} k_n(t, s) f(s) \ d\mu(s)$$

3. Note that each $T_n$ is compact because ... Well, I am actually having some trouble thinking of a good justification for this, and would appreciate some help on it. I also feel that it's somehow wrong.

4. The norm-limit of compact operators is compact, which implies that $T$ is compact.

Approximating the kernel by simple functions is not good enough. Let $\{e_j\}$ be an orthonormal basis for $L^{2} (X)$. Then the functions $(x,y) \to e_j (x) e_k (y)$ form an orthonormal basis for $L^{2} (X \times X)$. [ The completeness of this orthonormal family is proved by an application of Fubini's Theorem]. Hence the $L^{2}$ kernal k has an expansion in terms of this basis and the partial sums are finite sums of terms of the type $e_j (x) e_k (y)$. Now it is quite obvious that $T_n$'s defined by this approximating sequence have finite dimensional range (spanned by a finite number of $e_n$'s).
$T_n$ is a bounded operator and $\dim im(T_n) < \infty$, hence $T_n$ is compact.
• I was actually having trouble justifying that $\mathrm{dim} (\mathrm{Im} (T_n)) < \infty$. Why is this so? – Anton Xue Mar 7 '18 at 6:47
• because $T_nf$ is always a simple function equipped with characterization functions. – Lin Xuelei Oct 20 '18 at 7:51