Subset of a Banach space If $X$ is the space of all bounded continuous Real valued functions defined on $\mathbb{R}$ . For every $f$ belonging to $X$  define
$$
\|f\|_\infty=\{\sup|f(x)| ,x \text{ belongs to } \mathbb{R}\}.
$$
I have proved that $X$ is a Banach space. The question is: $Y=\{f \text{ belongs to } X , f \text{ is differentiable on } \mathbb{R}\}$, is $Y$ a closed subspace of $X$?
 A: The conclusion is false. Let $$f(x)=\begin{cases}
(1-x^2)^2(1-|x|) &|x|\le 1\\
0 & |x|>1
\end{cases}$$ Then $f$ is continuous, bounded and  not differentiable at $x=0.$ Indeed $f'_+(0)= -1$ and
$f'_-(0)=1.$ By the Weierstrass theorem there is a sequence of polynomials $p_n(x)$ such that $p_n(x)\rightrightarrows 1-|x|$ for $|x|\le 1.$ Let $$f_n(x)=\begin{cases} (1-x^2)^2 p_n(x) & |x|\le 1\\
0 & |x|>1
\end{cases}$$
Then $f_n$ is differentiable for $x\neq \pm 1.$ On the other hand
$$(f_n)'_+(1)=0=(f_n)'_-(1)=0,\qquad (f_n)'_+(-1)=0=(f_n)'_-(-1)=0$$
Hence $f_n$ is differentiable. Moreover $f_n\rightrightarrows f.$
A: Elaborating on the great answer of Ryszard Szwarc. The convergence of sequences with the norm you have defined is called "uniform convergence" and in Ryszard answer's is denoted by the double arrow.
Then, to show that your set $Y$ is not closed we can show that $X-Y$ is open. Ryszard has shown that there exists a function $f\in X-Y$ such that every open neighbourhood contains elements of $X$ ($f_n$ for some $n$). So $X-Y$ cannot be open, and $Y$ is not closed.
