Given a map $E:L^\infty\rightarrow C^1(\overline{\Omega}), g_n\rightarrow g$ in $L^\infty(\Omega)$. $u_n=E(g_n), n\in \mathbb{N}$ and u=E(g) and show that every time we take a subsection we can find a subsubsection that converge to $u \in C^1(\overline{\Omega})$then they say the function is continuous, why?


This is because of the following result:

Lemma: Let $X$ be a topological space and let $x_n, x \in X$. Then $x_n \to x$ if and only if every subsequence $x_{n_k}$ of the sequence $x_n$ has a further subsequence $x_{n_{k_j}}$ such that $x_{n_{k_j}} \to x$ as $j \to \infty$.

To prove the non-trivial direction in this lemma, you assume that $x_n \not \to x$ and use this to construct a subsequence that doesn't have any subsequence converging to $x$.

Continuity of your map $E$ then follows since for maps between metric spaces sequential continuity and continuity coincide so that it is enough to check that if $g_n \to g$ then $E(g_n) \to E(g)$. This is precisely what is provided by your hypotheses and the lemma.

  • $\begingroup$ Can you give me the proof of the lemma? I meant with details $\endgroup$ – Mono May 24 at 12:06
  • $\begingroup$ The lemma is really very easy to prove. You should try to do it yourself. For the non-trivial direction, if $x_n \not \to x$ then there is an open neighbourhood $u$ of $x$ such that there are infinitely many $n$ with $x_n \not \in U$. Use this to write down a subsequence that has no further subsequence that converges to $x$. $\endgroup$ – Rhys Steele May 24 at 13:10

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