Is the derivate on a closed subspace of $C^1[0,1]$ is a continuous linear map? I'm trying to show that $D:(X, \|\cdot\|_\infty) \rightarrow C[0,1]$  is a continuous map. $D$ is the differential operator and  $X$ is a closed (proper) subset of $C^1[0,1]$.
The fact that $X$ is closed in $C^1[0,1]$ must be important in the proof because otherwise this result is obviously false. However, I don't know how to use this fact.
I need this result to apply Arzela-Ascoli theorem to show that unit ball of X is compact and then conclude that $X$ is finite dimensional.
Does anyone know how to tackle this problem ?
 A: First, note that, $X$ is closed in $(C^1[0,1],∥⋅∥_{C^1})$, where, $∥f∥_{C^1}:=∥f∥_{\infty}+∥Df∥_{\infty}$. 
Indeed, suppose $(f_n)$ in $X$, such that $f_n \rightarrow f$ and $Df_n \rightarrow g$, uniformly. Then $f,g \in C[0,1]$ and, by fundamental theorem of calculus,
$$f_n(x)= f_n(0) + \int_0^x Df_n(t)dt.$$
So,
$$f(x)= f(0) + \int_0^x g(t)dt.$$
Thus, by fundamental theorem of calculus, $f\in C^1[0,1]$ and $Df=g$. This prove that $X$ is closed in $(C^1[0,1],∥⋅∥_{C^1})$.
Now, write $Id:(X,∥⋅∥_{C^11})→(X,∥⋅∥_{\infty})$. Note that $Id$ is a homeomorphism, By open mapping theorem. Thus, exist $c>0$, such that, for each $f\in X$,
$$\|Df\|_{\infty} \leq \|f\|_{C^1}\leq c\|f\|_{\infty},$$
that is, $D:(X,\| \cdot\|_{\infty}) \longrightarrow C[0,1]$ is a continuous map.
A: If $D$ is intended to be differentiation and if you really mean to use the infinity-norm (the supremum of the values of the function) on the domain, then $D$ isn't continuous.  You can have functions in $C^1$ that are very small in the infinity-norm but whose derivatives get very big --- a small function that wiggles a lot, like $\frac1n\sin(n^{100}x)$.
Perhaps the norm on the domain of $D$ should have been a more typical norm on $C^1$, taking into account not only the magnitude of the function but also the magnitude of its derivative.  Then $D$ would be continuous, and the proof of that would be very easy.
A: Take $X = C^1[0,1]$ and let $f_n \in X$ be $f_n(x) = \frac{1}{\sqrt{n}}x^n$. Then $\|f_n\|_{\infty} = \frac{1}{\sqrt{n}}$, but $\|Df_n\|_{\infty} = \sqrt{n}$. Hence $D$ is not continuous.  
