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.
- $T(B)$ is equibounded
$|Tu(x)|=|\int_0^1 \sin(x^2+y^2) u(y) dy| \leq 1\cdot ||u||_{L^2}^2 \leq M_b$ - 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?
Best Answer
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.