I'm trying to work through how Rellich compactness theorem is used at the start of this proof to extract the subsequence

  • Given an sequence $\{y_n\}^{\infty} _{n=1} \in H_A(D^*)/ \mathbb{R}$ that is bounded in the energy norm over $D^*$, you can extract a subsequence that converges in $H^1(D)$ to an element of $H_A(D)/ \mathbb{R}$.

The start of the proof was given as: Use Poincare Inequality along with Rellich compactness theorem to extract a convergent subsequence in $L^2(D^*)$.

Some notes on terminology:

  • $D \subset D^* \subset \mathbb{R}^n$ (for $n=2,3$),

  • the energy norm is $||x||^2 _{E(D)} = \int_{D} A \nabla x \cdot \nabla x \mbox{ } d \vec{x}$,

  • $A$ is a $n \times n$ symmetric positive definite matrix,

  • $H_A(D)$ is the space of functions in $H^1(D)$ that are A harmonic on $D$

My attempt so far is: Consider a sequence $\{y_n\}^{\infty} _{n=1} \in H_A(D^*)/ \mathbb{R}$ that is bounded in the energy norm over $D^*$. This means there exists some positive M such that

$||y_n||_{E(D^*)} =\int_{D^*} A \nabla y_n \cdot \nabla y_n \leq M$.

I was thinking that if I can show that the sequence is bounded in $H_0 ^1( D^*)$ then I can use Rellich compactness Theorem to extract a convergent subsequence in $L^2(D^*)$ (?).

I tried to follow what I saw here first but I'm not sure this is correct because I think I need to use the quotient space.

Next I tried to follow what I found here in the second answer. I took $y_0 \in H_A(D^*)/ \mathbb{R} \subset H^1(D^*)$ to be the average value of $y$ in $D^*$. This, I think, means that $y_n - y_0 \in H^1 _0(D^*)$ (?). Then

$||y_n - y_0||^2_{H^1 _0(D^*)}$ $ = \int_{D^*} (y_n - y_0)^2 d \vec{x} + \int_{D^*} \nabla(y_n - y_0)^2 d \vec{x}$ $\leq C \int_{D^*} \nabla(y_n)^2 \mbox{ } d \vec{x} +\int_{D^*} \nabla(y_n - y_0)^2 \mbox{ } d \vec{x} $

(last inequality is by alternate version of Poincare Inequality from this source)

$= (C+1) \int_{D^*} \nabla y_n \cdot \nabla y_n \mbox{ } d \vec{x} -\int_{D^*} 2 \nabla y_0 \cdot \nabla y_n + \nabla y_0 \cdot \nabla y_0 \mbox{ } d \vec{x}$

But here is where I get confused. I was hoping to be able to use the boundedness of the energy norm in this line of reasoning but without the $A$ matrix involved in the integrals I'm not sure how I can.

Once the above bounded argument is fixed, next we can say that by Rellich compactness theorem there exists a subsequence $\{u_{n_k}\} = \{Ty_{n_k}\}$ that converges in $L^2(D^*)$ (I think?). Then since the subsequence converges it is Cauchy in $L^2(D^*)$.

Is this the correct way to use Rellich compactness theorem in the start of this proof? Any input is greatly appreciated. Thank you greatly in advance!

If you're interested, I'm referencing this paper.

  • 1
    $\begingroup$ This has nothing to do with Rellich theorem. You can use that $H_A(D^*)/\mathbb R$ is a Hilbert space, then use weak convergence on that Hilbert space. $\endgroup$
    – daw
    Commented Feb 9 at 8:34
  • $\begingroup$ @daw Thank you for your comment! I think we needed more than weak convergence for what they do next in the paper. In the proof listed they reference Rellich compactness directly so I was hoping to understand more of the details behind what they meant. $\endgroup$
    – k12345
    Commented Feb 9 at 18:43
  • 1
    $\begingroup$ The following will be useful regarding the matrix: math.stackexchange.com/questions/1115480/… $\endgroup$
    – JCO
    Commented Feb 9 at 22:02

1 Answer 1


For the bounded sequence $(\tilde{y}_n)$ in $H_A(D^{*})/\mathbb{R}$, we obtain a bounded sequence $(y_n)$ with zero mean: for every $n$, subtract $\tilde{y}_n$ from its mean (see next). Since $y_n$ and $\tilde{y_n}$ differ by a constant, they are in the same equivalence class in $H_{A}(D^{*})/\mathbb{R}$ (i.e., $y_n = \tilde{y}_n$ in $H_{A}(D^{*})/\mathbb{R}$).

$y_n = \tilde{y}_n - \frac{1}{|D^{*}|}\int_{D^{*}}\tilde{y}_n dx$

Now, we have

$\|y_n\|^2+\|\nabla y_n\|^2 \leq c_1 \|\nabla y_n\|^2 + c_2 \|A \nabla y_n\|^2 \leq c_3 \|A \nabla y_n\|^2 + c_2 \|A \nabla y_n\|^2\leq C \|A \nabla y_n\|^2 \leq M.$

In the inequalities above we used the coercivity property of A (link in my comment) and Poincare's inequality for functions in $H^1$ with zero mean ($H_{0}^{1}$ has no role: we will use the Theorem of Rellich-Kondrachov for $H^1$)

Thus, $y_n$ is bounded in $H^1$ and by the compactness of the embedding $H^1 \subset L^2$, there exists a subsequence of $(y_n)$ that converges strongly in $L^2$.

  • $\begingroup$ Thank you very much for the response! I think I see what you mean but I have a few questions before I award the bounty: (1.) How do we obtain the bounded sequence $(y_n)$ with zero mean? Can you explain more about that trick please and maybe what role the quotient space plays a role in it? (2.) Thanks for the link! So the coercivity property of A also works for the gradient of $y_n$ i.e $ |\nabla y|^2 \leq \frac{1}{c^*}<A \nabla y,\nabla y>$ ($c^*$ is from the coercivity)? (3.) And in the second inequality again we use Poincare and the term next to $c_2$ should be $||A \nabla y_n||$? $\endgroup$
    – k12345
    Commented Feb 11 at 19:54
  • 1
    $\begingroup$ @k12345 I have fixed the typo that you noticed (thanks) and added the calculation of $y_n$. I also included explicit mention of the relation of this with the quotient space $H_{A}/R$. Thank you for bringing this problem. $\endgroup$
    – JCO
    Commented Feb 11 at 21:11
  • 1
    $\begingroup$ @ k12345 Regarding (2), only the coercitivity is used in the second inequality, and yes this condition holds for the gradient too as you mentioned. The constant $c_3$ has the Poincare's constant $c_1$ and the coercitivity constant $c^{*}$ in it. The constant $c_2$ has only the coercitivity constant in it. Let me know in case I left something without proper explanation. $\endgroup$
    – JCO
    Commented Feb 12 at 1:11
  • $\begingroup$ Thank you so much for such a detailed and helpful answer! I truly appreciate you taking the time to explain all of this and answer all of my questions. This was very helpful for me. $\endgroup$
    – k12345
    Commented Feb 12 at 5:22

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