Find the closure of $C^{1}[-1, 1]$ in the space ${(\scr{B}} [-1, 1],||•||_\infty)$

Solution 1:

The closure is $C[-1,1]$. Any function in the closure of $C^{1}[-1,1]$ is continuous because uniform limits of continuous functions are continuous. On the other hand any continuous fucntion can be approximate uniformly by a sequence of polynomials and polynomials are in $C^{1}[-1,1]$.

Solution 2:

$C^1[-1, 1] \subset C[-1, 1]$(closed set)

Hence, $\overline{C^1[-1, 1] } \subseteq C[-1, 1]$

Now we prove, $C[-1, 1] \subseteq{\overline{ C^{1} [-1, 1]}}$

Choose, $f\in C[-1, 1]$ then by Weierstrass approximation theorem, there exists a sequence of polynomial $(p_n)_{n \in \mathbb{N}} \subset {\scr{P[-1, 1]}}$ such that

$(p_n) \to f $ uniformly on $[-1, 1]$

In other words, $|| p_n -f||_{\infty} \to 0 $ as $n \to \infty $ in $(C[-1, 1],||•||_\infty)$

Hence, $\overline{C^1[-1, 1] } = C[-1, 1]$