I am confused by the following approach that is presented in a proof in a paper:
Suppose that $D$ is the domain, $H$ is a Hilbert space that embeds continuously into $L^{\infty}(D)$, and $H_0^1(D)$ is the Sobolev space we're familiar with. Let $q \in H$ and consider the mapping $u: q\in H \rightarrow u(q) \in H_0^1$. The paper is trying to show that $u$ is a continuous mapping via a subsequence argument that I will show below. Assume that we have access to an inequality (*) that is derived from another paper. The inequality says:
(*) $\|u(q') - u(q)\|_{H_0^1} \le C \|q'-q\|_{L^{\infty}(D)} $.
The proof goes as follows. The idea is that for $q^n \rightarrow q$ in $H$, we want to show that $u(q^n) \rightarrow u(q)$ to prove continuity.
1) Because $H$ continuously embeds into $L^{\infty}(D)$, there is a subsequence $q^{n_k}$ such that $\|q^{n_k}-q\|_{L^{\infty}} \rightarrow 0$.
2) Consequently, using (*), we have that for this subsequence $q^{n_k}$, $\|u(q^{n_k}) - u(q)\|_{H_0^1} \rightarrow 0$.
The proof then says: because any subsequence of $u(q^{n_k})$ contains a subsequence that converges to $u(q)$, as a consequence, we have that $u(q^n) \rightarrow u(q)$.
I am confused by the part where "any subsequence of $u(q^{n_k})$ contains a subsequence that converges to $u(q)$". This is because if we go back to point 1), it says that there exists a specific subsequence $q^{n_k}$ with $\|q^{n_k}-q\|_{L^{\infty}} \rightarrow 0$. In other words, if we take a subsequence of $u(q^{n_k})$, how do we ensure that this subsequence contains elements of $q^{n_k}$ (that we obtained earlier) for which the $L^{\infty}$ error of $q^{n_k}$ goes to zero?
Am I making sense or could the proof had been stated differently?
The argument works the same if we start with an arbitrary subsequence of $(q^n)$. That has a subsequence such that … and so on.
But it's an unnecessarily roundabout argument. Since $H$ embeds continuously into $L^{\infty}(D)$, $q^n \to q$ in $H$ immediately implies $q^n \to q$ in $L^{\infty}(D)$, and then $(\ast)$ directly yields $u(q^n) \to u(q)$ in $H^1_0(D)$. There's no need to consider subsequences.