What role does the range of an operator play in showing the operator is compact?

Question

To show an operator is compact I understand you have to show the operator is the limit of finite rank operators. However the proof I have doesn't do this.

I have an operator $$k:C0[,\pi]) \to C([0, \pi])$$ given by $$(Ku)(x)=\int^\pi _0 (sin x+ cos t)u(t) dt$$ where $u \in C([0, \pi])$

I have the proof to show the operator is compact.

Firstly we show the operator is bounded and it follows that $||Ku|| \leq 2 \pi ||u||$. Hence the operator is bounded.

It says in my notes that $(Ku)(x)=\mathrm{a}\ \mathrm{sin}\ x + b$ for some constants a,b. How do we get this?

From this it follows that $\mathrm{Ran} K \subset \ \mathrm{span}\{\sin,1\}$

Which shows that the operator is compact.

Why do we have to look at the range of K? What role does this play in showing the operator is compact?

Answer 1

There is a theorem that so-called finite-rank operators between normed spaces are compact. It goes like this:

Theorem 8.1-4 [Kreyszig: Introductory Functional Analysis with Applications]

Let $X$ and $Y$ be normed spaces and $T:X\to Y$ a linear operator. Then if $T$ is bounded and dim$(T(X))<\infty$ the operator is $T$ is compact.

(Proof given in the text)

In your case $K$ is bounded and its range is spanned by 2 vectors, i.e. it's 2-dimensional, hence it is compact by the theorem.

For your first question, on why you can write $(Ku)(x) = a\sin x + b$, see the answer by @Tryss.

Answer 2

For your first question,

$$Ku(x) = \int_0^\pi (\sin(x)+\cos(t)) u(t) dt = \sin(x) \underbrace{\int_0^\pi u(t) dt}_a + \underbrace{\int_0^\pi \cos(t)u(t) dt }_b$$

For your second question, if the range of an operator is a finite dimensional vector space, then it is compact, because the unit ball of a finite vector space is compact

Answer 3

This is an integral operator with kernel $K:C([0,\pi])^2\to\mathbb R$ defined by $K(x,t)=\sin x+\cos t$. Since $$\|K\|_2^2=\int_0^\pi\int_0^\pi(\sin x+\cos t)^2\ \mathsf dx\ \mathsf dt =\pi^2<\infty,$$ $K$ is a Hilbert-Schmidt kernel and therefore the operator $k$ is compact.

(The proof of this result does use a sequence of finite-rank operators to approximate the given integral operator.)