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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?

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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.

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  • $\begingroup$ Since $a,b$ are depending on $u$, how can I be sure that the are bounded also when $u$ blows up? $\endgroup$ – VoB Jan 26 at 23:26
  • $\begingroup$ @VoB you can easily verify that $a$ and $b$ are finite...just try to figure out what they are. $\endgroup$ – Shashi Jan 26 at 23:32
  • $\begingroup$ Maybe you want to remark that one gets this observation using $$ \sin(x^2+y^2) = \cos(x^2)\sin(y^2) + \sin(x^2) \cos(y^2) $$ $\endgroup$ – Severin Schraven Jan 26 at 23:37
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    $\begingroup$ 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) $\endgroup$ – VoB Jan 31 at 18:51
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    $\begingroup$ 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 $Y$. $\endgroup$ – VoB Jan 31 at 19:26

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