Sobolev inequality in one dimension

Question

In Brezis's book, it states the Sobolev inequality in 1D

For any $u\in W^{1,p}(\mathbb{R}), 1\leq p\leq\infty$, there exists a constant C such that $||u||_{\infty}\leq C||u||_{W^{1,p}}$

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First, it proves the case when $u\in C_{c}^{\infty}(\mathbb{R})$. Then by the density argument, there exists $u_n\rightarrow u$ in $W^{1,p}$. It says that since $\{u_k\}$ is a Cauchy sequence in $L^{\infty},$ then $u_n\rightarrow u$ in $L^{\infty}$. I couldn't see why it is true

Answer 1

Once you know that for all $u \in C_c^\infty(\mathbb{R})$ we have that $\|u\|_\infty \leq C \|u\|_{W^{1,p}}$ it is clear that Cauchy sequences for $\|\cdot\|_{W^{1,p}}$ in $W^{1,p}(\mathbb{R}) \cap C_c^\infty(\mathbb{R})$ are also Cauchy for $\|\cdot\|_\infty$.

In particular, if $u_n \to u$ in $W^{1,p}(\mathbb{R})$, then $u_n$ is Cauchy for $\|\cdot \|_\infty$ and hence there is $v \in L^\infty(\mathbb{R})$ such that $u_n \to v$ in $L^\infty$ and in particular $u_n \to v$ a.e.

However, $u_n \to u$ in $W^{1,p}$ implies that $u_n \to u$ in $L^p$ and hence there is a subsequence $u_{n_k}$ such that $u_{n_k} \to u$ a.e. Since we must have $u_{n_k} \to v$ a.e. this gives that $u = v$ a.e. so $u_n \to u$ in $L^\infty$.