# Compactness of an integral operator from $L^2$ to $L^2$

I want to prove that

The operator (linear and bounded) $$T: L^2(0,1) \rightarrow L^2(0,1)$$, defined by: $$Tu(x)=\int_0^1\sin(x^2+y^2)u(y)dy$$, is compact.

Just by using theory, it's an Hilbert Schmidt operator, so it's compact. Indeed, the kernel $$k(x,y) = \sin(x^2+y^2) \in L^2([0,1] \times [0,1])$$.

I want to use a more direct approach, by using Ascoli-Arzelà.

First of all, I see that

$$|Tu(s)-Tu(t)|=| \int_0^1 [\sin(s^2+y^2)-\sin(t^2+y^2)]u(y)dy| < \varepsilon ||u||_{L^2}^2$$, since $$s \mapsto \sin(s^2+y^2)$$ is continuous on a compact set, then it's unformly continuous.

Then $$Tu(x)$$ is continuous, and then $$T: L^2([0,1]) \rightarrow C^{0}[0,1] \subset L^2([0,1])$$

To prove compactness, I take $$B \subset L^2(0,1)$$, with $$||u|| \leq M_b$$, for $$u \in B$$. I want to prove that $$T(B)$$ is relatively compact in $$C^0([0,1])$$ by using Ascoli-Arzelà.

1. $$T(B)$$ is equibounded $$|Tu(x)|=|\int_0^1 \sin(x^2+y^2) u(y) dy| \leq 1\cdot ||u||_{L^2}^2 \leq M_b$$
2. Now I show equicontinuity

Again, I compute

$$|Tu(s)-Tu(t)| = \int_0^1 [\sin(s^2+y^2)-\sin(t^2+y^2)]u(y)dy| \leq |s-t| ||u||_{L^2}^2$$

by MVT applied to $$s \mapsto \sin(s^2+y^2)$$ for $$s \in [0,1]$$.

So $$T(B)$$ is equilipschitz, thus equicontinuos.

Then, by Ascoli-Arzelà, $$T(B)$$ is relatively compact in $$C^0([0,1])$$. Now, since the immersion $$C^0 \hookrightarrow L^2[0,1]$$ is continuos, then $$T(B)$$ is relatively compact in $$L^2[0,1]$$.

Is my second approach okay? Or do I need to fix something?

Your approach is correct. But there is an easier way. Observe that $$Tu(x)=a\sin(x^2)+b\cos(x^2)$$ where $$a$$ and $$b$$ are constants depending on $$u$$. Thus, the range of $$T$$ is finite-dimensional and $$T$$ is compact.
• Since $a,b$ are depending on $u$, how can I be sure that the are bounded also when $u$ blows up? – VoB Jan 26 at 23:26
• @VoB you can easily verify that $a$ and $b$ are finite...just try to figure out what they are. – Shashi Jan 26 at 23:32
• Maybe you want to remark that one gets this observation using $$\sin(x^2+y^2) = \cos(x^2)\sin(y^2) + \sin(x^2) \cos(y^2)$$ – Severin Schraven Jan 26 at 23:37
• To show boundedness of the operator $T$ I need to prove that $||(Tu)(x)||_{L^2} \leq C ||u||_{L^2}$, for some $C >0$. I have to compute a double integral, (in dy dx) – VoB Jan 31 at 18:51
• This come from the fact that $T:L^2(0,1) \rightarrow L^2(0,1)$. If you do in your way, you're assuming that $T:L^2(0,1) \rightarrow \mathbb{R}$, which is not the case. In the boundedness inequality, for $T:X \rightarrow Y$, you always have to prove $||Tu||_{Y} \leq C ||u||_{X}$, so it depends on the norm you have on $X$ and $Y$. – VoB Jan 31 at 19:26