Sequence converges if every subsequence has a convergent subsequence with unique limit

Question

I Want to Show

Let $\lambda\in\Lambda:=\{x\in\mathbb{R}\,|\,x>0\}$ and let $(u_{\lambda})\in H^{1}(\Omega)$, $\Omega\subset\mathbb{R}^{N}$ bounded and smooth, then $u_{\lambda}\rightharpoonup u$ in $H^{1}(\Omega)$ if for every sequence $(u_{\lambda_{n}})_{n\in\mathbb{N}}$ there exists a subsequence such that $u_{\lambda_{n_{k}}}\rightharpoonup u$ with $u$ unique for every convergent subsequence.

Proof

We prove the contrapositive, that is, there exists a sequence $(u_{\lambda_{n}})$ such that for every subsequence $u_{\lambda_{n_{k}}}\not\rightharpoonup u$ if $u_{\lambda}\not\rightharpoonup u$.

Assume $u_{\lambda}\not\rightharpoonup u$ in $H^{1}(\Omega)$. We construct our sequence $(u_{\lambda_{n}})$ as follows. For every $n\in\mathbb{N}$ there exists $\lambda\in\Lambda$ such that $\lambda=n$, then for every $n\in\mathbb{N}$ set $\lambda_{n}:=\lambda=n$. So there exists $\varepsilon>0$ such that for all $N\in\mathbb{N}$ there exists $\lambda_{n}\geq N$ and $\varphi\in H^{1^{*}}(\Omega)$ with, \begin{align} |\varphi(u_{\lambda_{n}})-\varphi(u)|\geq\varepsilon. \end{align} That is, there are all but finitely many $u_{\lambda_{n}}$ such that $\varphi(u_{\lambda_{n}})$ is within distance $\varepsilon$ of $\varphi(u)$. Hence for every subsequence of $(u_{\lambda_{n}})$ we have $u_{\lambda_{n_{k}}}\not\rightharpoonup u$.

My Question

I am not confident with the contrapositive statement. I know that I have not accounted for the uniqueness of $u$ in the contrapositive statement and this is because I do not know how to. Clearly the opposite of unique is non-unique but does this need to be specified in my contrapositive statement?

Answer 1

You got the negation of the weak convergence statement wrong. If $u_\lambda\not\rightharpoonup u$ then there is $\phi \in H^1(\Omega)^*$ such that $\phi( u-u_{\lambda}) \not\to0$. Then there are $\epsilon>0$ and $(\lambda_n)$ such that $$ |\phi( u-u_{\lambda_n})| > \epsilon $$ for all $n$. This is your subsequence: no subsequence of $(u_{\lambda_n})$ will converge weakly to $u$.

This proof accounts also for possible non-uniqueness of weak limits: If the sequence has two subsequences with different weak limits, then the whole sequence is not converging to $u$. So this situation is included in the reasoning.

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Note, that the convergence statement (each subsequence has converging subsequence ...) has nothing to do with $H^1$, weak convergence, etc. In fact, it is true in topological spaces.

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In the OP, the negation of $u_n\rightharpoonup u$ is wrong. By definition, $u_n\rightharpoonup u$ if and only if for all $\phi$ we have $\phi(u_n-u)\to0$. This is equivalent to: for all $\phi,\epsilon>0$ there is $N$ such that $|\phi(u_n-u)|<\epsilon$ for all $n>N$. So the negation is: There is $\phi$ and $\epsilon>0$ such that for all $N$ there is $n>N$ with $|\phi(u_n-u)|\ge\epsilon$. Now you can construct the subsequence.

Answer 2

Assume $u_{\lambda}\not\rightharpoonup u$ in $H^{1}(\Omega)$, with $\lambda\in\Lambda:=\{x\in\mathbb{R}\,|\,x>0\}$.

We first construct a sequence $(u_{\lambda_{n}})$ which does not converge weakly. Consider all $\lambda\in\Lambda\cap\mathbb{N}$. Then by assumption $u_{\lambda}\not\rightharpoonup u$, so by setting $\lambda_{n}:=\lambda=n$, for every $n\in\mathbb{N}$, we then have by definition that $\exists\varphi\in(H^{1}(\Omega))^{*}$ and $\varepsilon>0$ such that $\forall N\in\mathbb{N}$ there exists $\lambda_{n}\geq N$ with, \begin{align} |\varphi(u_{\lambda_{n}})-\varphi(u)|\geq\varepsilon. \end{align}

So for every subsequence $(u_{\lambda_{n_{k}}})$ the construction of $(u_{\lambda_{n}})$ ensures that there exists $\varphi\in(H^{1}(\Omega))^{*}$ and $\varepsilon>0$ such that $\forall N_{k}\in\mathbb{N}$ there exists $\lambda_{n_{k}}\geq N_{k}$ with, \begin{align} |\varphi(u_{\lambda_{n_{k}}})-\varphi(u)|\geq\varepsilon. \end{align}

Hence every subsequence $(u_{\lambda_{n_{k}}})$ of $(u_{\lambda_{n}})$ does not weakly converge to $u$.