0

Let $T\colon H^2([a,b])\to L^2([a,b])$ be some linear operator.

Show that it is densely defined and has compact resolvent (i.e. $T^{-1}$ has dense range and is compact).

Hint: Use that $H^2([a,b])$ is compactly embedded in $L^2([a,b])$.

I do not know exactly if I got this task right: Is it meant in the sense of an equivalence, that is: it is enough to prove that $T^{-1}$ has dense range and is compact to prove that $T$ is densely defined and has compact resolvent? Or is the statement in parantheses only an implication?

I am a bit confused.

Anyway, the given hint reminds me on Embedding compact iff resolvent compact

Salamo
  • 1,094
  • 1
    I don't think you mean $T : H^2 \rightarrow L^2$. I think you mean $T : H^2 \subset L^2 \rightarrow L^2$. Please check on that. – Disintegrating By Parts May 12 '17 at 17:25
  • On my exercise sheet it is indeed written $T\colon H^2\to L^2$ but I am quite sure what is really meant is $T\colon H^2\subset L^2\to L^2$. – Salamo May 12 '17 at 18:03
  • It's critical to distinguish between these, because the norm being used in the range depends on the underlying space. Writing $T : H^2 \rightarrow L^2$ comes with an implied $H^2$ norm on the domain, whereas $T : H^2 \subset L^2 \rightarrow L^2$ comes with an implied $L^2$ norm on the domain. – Disintegrating By Parts May 12 '17 at 18:35
  • Then maybe I just choose the one with which the Statement maybe works. :-) – Salamo May 12 '17 at 18:39
  • @DisintegratingByParts In the tutorial it was just said that the Statement follows by the fact that if $L^{-1}$ exists and is bounded, the compactness follows by the compact embedding. To be honest, I did not understand that and in fact, this seems too easy to me. Maybe you can give me a hint? – Salamo May 13 '17 at 14:31
  • I think the only way to interpret this to get compactness of the resolvent is to assume that $|T^{-1}f|{H^2} \le C|f|{L^2}$. That would do it because $T^{-1}$ would map a bounded subset in $L^2$ into a bounded subset in $H^2$, which would be compact in $L^2$. So I'd say that's how it must be interpreted in order to get the desired result. – Disintegrating By Parts May 13 '17 at 16:41

0 Answers0