Compact Integral operator?

Question

I have stumbled upon an interesting linear operator. I am not sure if it is compact: $$ T: L^2([0, 1]) \rightarrow L^2([0, 1]), \quad (Tf)(x) := \int^1_0 (x-y)^2 f(y)~\mathrm{d}y. $$

I tried to use the following argument: Let $(f_k)_{k \in \mathbb{N}} \subseteq L^2([0, 1])$ be bounded and let $(f_{n_k})_{k \in \mathbb{N}} \subseteq L^2([0, 1])$ be a subsequence with weak limit $f \in L^2([0, 1])$. Then: $$ \lVert Tf_{n_k} - Tf \rVert_{L^2([0, 1])}^2 = \int_0^1 \left \lvert \int^1_0 (x-y)^2 (f_{n_k}(y) - f(y))~\mathrm{d}y \right \rvert^2 ~\mathrm{d}x \leq \sup_{x \in [0, 1]} \left \lvert \int^1_0 (x-y)^2 (f_{n_k}(y) - f(y))~\mathrm{d}y \right \rvert^2 $$ I can still prove that this $\sup$ is attained on some $x_{n_k} \in [0, 1]$ but this does not help me because of the dependancy on $n_k$. Otherwise I would have used weak convergence at that point.

Is there an easier way or this is operator even compact?

Answer 1

It's a finite rank operator and thus compact: $$(Tf)(x)=A\,x^2-B\,x+C,$$ where $A=\int^1_0f(y)\,dy, B=2\int^1_0y\,f(y)\,dy, C=\int^1_0y^2\,f(y)\,dy$, so the image of a bounded set is a bounded subset of a 3-dimensional subspace of $L^2([0, 1])$.

Answer 2

If $k(x,y)=a(x)b(y)$, where $a$ and $b$ are continuous functions on $[0,1]$, then $$ \int_0^1 k(x,y) f(x)\, dx = \left(\int_0^1 a(x) f(x)\, dx\right) b(y) = \langle a,f\rangle b(y). $$ Therefore the integral operator $T_k$, given by the integral kernel $k$, satisfies $T_k(f) = \langle a,f\rangle b$, and, as a consequence, $T_k$ is seen to be a rank one operator. If instead, $$ k(x,y)=\sum_{i=1}^na_i(x)b_i(y), \tag{1} $$ where $a_i$ and $b_i$ are again continuous functions on $[0,1]$, the corresponding integral operator satisfies $$ T_k(f) = \sum_{i=1}^n \langle a_i,f\rangle b_i, \quad \forall f\in L^2[0,1], $$ so the rank of $T_k$ is at most $n$, hence finite.

If we next assume that $k$ is a uniform limit of functions $k_n$ of the form (1), it is not hard to see that $T_{k_n}\to T_k$, in operator norm, so $T_k$ is a compact operator since it is the limit of finite rank operators.

The Stone-Weierstrass Theorem may be easily invoked to show that the subset of $C([0,1]\times[0,1])$ formed by all functions of the form (1) is dense, so the conclusion of the paragraph above holds for any continuous $k$. This proves:

Theorem. Any integral operator on $L^2[0,1]$ with a continuous integral kernel is compact.