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Let $\Sigma$ a smooth surface that separates $\Omega_1$ and $\Omega_2$ (open and bounded sets) and let $(q^n_1)_{n\in\mathbb{N}}$ and $(q^n_2)_{n\in\mathbb{N}}$ sequences in $H^1(\Omega_1)$ and $H^1(\Omega_2)$, respectively. Suppose that $(q^n_1)_{n\in\mathbb{N}}$ converge to $q_1\in H^1(\Omega_1)$ and $(q^n_2)_{n\in\mathbb{N}}$ converge to $q_2\in H^1(\Omega_2)$. Then the trace over $\Sigma$ of each $q_1^n$ and $q_2^n$ is well defined, that is, for all $n\in\mathbb{N}$ $q_1^n|_\Sigma$ and $q_2^n|_\Sigma$ are well defined.

Prove the following:

if ($\forall n\in\mathbb{N}$) $q_1^n|_\Sigma-q_2^n|_\Sigma$ is constant, then $q_1|_\Sigma-q_2|_\Sigma$ is constant.

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My solution:

using the continuity of trace operator on $H^1(\Omega_1)$ and $H^1(\Omega_2)$ it is clear that:

$q_1^n|_\Sigma\rightarrow q_1|_\Sigma$ in $H^{1/2}(\Sigma)$

and

$q_2^n|_\Sigma\rightarrow q_1|_\Sigma$ in $H^{1/2}(\Sigma)$

and then, using the Cauchy-Schwarz inequality:

$\|q_1^n|_\Sigma-q_2^n|_\Sigma -( q_1|_\Sigma-q_2|_\Sigma)\|_{H^{1/2}(\Sigma)}\leq \|q_1^n|_\Sigma-q_1|_\Sigma\|_{H^{1/2}(\Sigma)}+\|q_2^n|_\Sigma-q_2|_\Sigma\|_{H^{1/2}(\Sigma)}\rightarrow 0$

$\Rightarrow (q_1^n|_\Sigma-q_2^n|_\Sigma) \rightarrow (q_1|_\Sigma-q_2|_\Sigma)$ in $H^1(\Sigma)$.

but, why $\boldsymbol{ (q_1|_\Sigma-q_2|_\Sigma)}$ is constant? I suppose that it is very clear, but I can't see it.

yemino
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1 Answers1

1

Let $C_n=q_1^n|_\Sigma-q_2^n|_\Sigma$ be the the sequence of constant functions, which as you alread have noted, converge to $q_1|_\Sigma-q_2|_\Sigma$ in $H^{1/2}(\Sigma)$.

Therefore, $C_n \to q_1|_\Sigma-q_2|_\Sigma$ in $L^2(\Sigma)$, or equivalently, $$\int_\Sigma |C_n-(q_1|_\Sigma-q_2|_\Sigma)|^2d\Sigma\to 0.\tag{1}$$

There are some ways to prove now that $q_1|_\Sigma-q_2|_\Sigma$ is constant, for example, from $(1)$, we can conclude that there is a subsequence $C_{n_k}$ of $C_n$, such that $C_{n_k}\to q_1|_\Sigma-q_2|_\Sigma$ a.e..

However, $C_{n_k}$ must converge a.e. to a constant $C$ (because it is a sequence of constant functions), thus, $q_1|_\Sigma-q_2|_\Sigma=C$ is constant.

Tomás
  • 22,559