With $V=L^2(0,T;H^1(\Omega))$, let $A:V \to V^*$ with $$\langle Au,v \rangle = \int_0^T \int_{\Omega} \nabla u(t) \cdot \nabla v(t).$$

I want to show that $A$ is a compact operator.

So, one way to do this is, I need to show that if $v_n \in V$ with $\lVert v_n \rVert_V = 1$, then $Av_n$ has a Cauchy subsequence.

Well, $v_n \rightharpoonup v$ in $V$ obviously but this doesn't give me $Av_n \to Av$ which is what I was hoping for. Any other way I can show compactness? Thanks

  • $\begingroup$ What makes you convinced $A$ is compact? $\endgroup$
    – Sam
    Jul 11 '13 at 22:09
  • $\begingroup$ @Sam Nothing, I just hope it is.. $\endgroup$
    – matt.w
    Jul 11 '13 at 22:10
  • 2
    $\begingroup$ I don't think this can be the case. If you consider the subspace of $L^2(0,T;H^1(\Omega))$ consisting of those elements with zero mean, $\int_\Omega u(t) = 0$, for almost every $t\in [0,T]$, then $\langle Au,v\rangle$ defines an inner product. So if $A$ where compact, then this subspace would have compact unit-ball, hence would be finite-dimensional. But it isn't. $\endgroup$
    – Sam
    Jul 11 '13 at 22:22
  • $\begingroup$ I do agree with @Sam, in fact, $A$ is inejective and I think that it is also bijective too. $\endgroup$
    – Tomás
    Jul 11 '13 at 23:30

Given a Hilbert space $H$, we can define the duality map $T:H\to H^*$ by $Tx(y)=\langle y,x\rangle$. This map is a bijection between $H$ and $H^*$; in fact, one sometimes uses it to "identify" $H$ with its dual, thus declaring $H$ to be the identity map. Of course, $T$ is not compact.

Let $H=L^2(0,T,K)$ where $K$ is also a Hilbert space. Then $T$ is defined by $$Tx(y)=\int_0^T \langle y(t),x(t)\rangle_K\,dt\tag1$$ Following Sam's suggestion, specialize the above to $K\subseteq H^1 (\Omega) $ being the set of functions with zero mean, equipped with $\langle u,v\rangle_K=\int_\Omega \nabla u\cdot \nabla v$. Then the duality map $T$ is exactly your operator; as stated in the first paragraph, it is not compact.

  • $\begingroup$ Thank you. But does this not contradict parabolic PDE theory where we use the eigenfunctions of the Laplacians to form o.n basis? I know the solution operator is compact in that case, so maybe it is not required for the Laplacian to be. $\endgroup$
    – matt.w
    Jul 14 '13 at 17:17
  • 1
    $\begingroup$ @matt.w The solution operator $\Delta^{-1}$ is compact from $L^2$ to $L^2$. It is not compact from $H^1$ to $(H^1)^*$. $\endgroup$
    – 40 votes
    Jul 14 '13 at 17:23
  • $\begingroup$ @matt.w Just a little addition, the rule of thumb is that differential operator is most of the time not compact, while integral operator is sometimes compact, $\Delta^{-1}$ can be loosely viewed as an integral operator. $\endgroup$
    – Shuhao Cao
    Jul 16 '13 at 18:39
  • $\begingroup$ Actually, I should have "$(H^1)^*$ to $H^1$" in the comment above. Or the Laplacian going in the opposite direction. They are both pretty close to being isomorphisms, hence very far from being compact. $\endgroup$
    – 40 votes
    Jul 16 '13 at 18:54

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