# Sequentially continuity

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?

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.
• 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 $x$. – Rhys Steele May 24 at 13:10