Find the closure of $C^{1}[-1, 1]$ in the space ${(\scr{B}} [-1, 1],||•||_\infty)$ Find the closure of $C^{1}[-1, 1]$ in the space ${(\scr{B}}[-1, 1],||•||_\infty)$
\begin{align} C^{1}[-1, 1]= &\{f:[-1,1] \to \mathbb{R}&: f' \text {  exists and continuous }\} \end{align}
\begin{align} {\scr{B}}[-1, 1]= &\{f:[-1,1] \to \mathbb{R}&: f \text { is bounded }\} \end{align}
$C^{1}[-1, 1]$ is not closed in the space $({\scr{B}}[-1, ,1],||•||_\infty)$.
Consider $C^{1}[-1, 1]$
$(f_n)_{n\in \mathbb{N}} \subset C^{1}[-1,1]$ defined by $f_n(t) =\sqrt{t^2 +\frac{1}{n^2}}$ $\{\forall t\in [-1, 1]\}$
Then, $(f_n) \to f $ in $({\scr{B}}[-1, 1],||•||_\infty)$.
Where, $f:[-1,1]\to \mathbb{R}$ defined by $f(t) =|t|$
\begin{align}||f_n - f||_\infty &=sup\{|f_n(t)-f(t)|: t\in [-1,1] \}\\
&=\frac{1}{n} \to 0 \space as\space  n\to \infty
\end{align}
Hence, $(f_n) \to f $ in $({\scr{B}}[-1, 1],||•||_\infty)$
But here, $f\notin C^1[-1,1]$
Hence, $C^1[-1, 1]$ is not closed in $({\scr{B}}[-1, 1],||•||_\infty)$.
I know $C^1[-1, 1]  \subset C[-1, 1]$(closed set)
My question : What is the smallest closed set containing $C^1[-1, 1]$ in the space $({\scr{B}}[-1, 1],||•||_\infty)$ ?

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 $C^{1}[-1,1]$.
