Is the space $C^1[a,b]$ with its usual 'sup-norm ' topology is separable?

Question

Is the space $C^1[a,b]$ with its usual 'sup-norm ' topology is separable

My attempt : i know $C[a,b] $ is separable because polynomials are dense in $C[0,1]$

But Here im confused with $C^1[a,b]$ ?

Answer 1

If $f$ is continuously differentiable and $\epsilon >0$ then there exist a polynomial $q$ such that $\|f'-q\|_{\infty} <\epsilon$. Since $f(x)=f(0)+\int_0^{x} f'(t)dt$ it follows that $\|f-p\|_{\infty} \leq \epsilon$ where $p(x)=f(0)+\int_0^{x} q(t) dt$. Thus $\|f-p\|_{\infty} \leq \epsilon$ and $\|f-p\|_{\infty} \leq \epsilon$. Note that $p$ is also a polynomial.Now aprroximate the coefficients of $p$ by rational numbers to see that polynomials with rational coefficients form a countable dense set.

Answer 2

$C^1[a,b]$ is the space of differentiable functions with a continuous derivative.

$C^1[a,b]$ is separable since the polynomials are dense in $C^1[a,b]$.

Firstly, note that polynomials are in $C^1[a,b]$. Secondly, note that $C^1[a,b]$ is a subset of $C[a,b]$, and that therefore if every function in $C[a,b]$ can be closely approximated with a polynomial then so can every function in $C^1[a,b]$.