TITLE: The closure of differentiable functions are the continuous functions in the interval $[0,1]$
QUESTION [5 upvotes]: Let $X=C([0,1])$ with the $||\cdot||_\infty$-norm. Show that $\overline{C^1([0,1])}=X$.

Every differentiable function is continuous. So $C^1([0,1])\subset X$. Now I have to show $D=\overline{C^1([0,1])}\subset X$. I also have to show: $X\subset D$. So every continuous function which is not differentiable is in $\overline{D}/\dot{D}$. 
Every continuous function f in a compact intervall is bounded, so $||f||_\infty<\infty$. $C^1([0,1])\subset X$, so $||f||_\infty<\infty \forall f\in C^1([0,1])$. So for the functions g in the closure of $C^1([0,1])$ we have $||g||_\infty<\infty$. How can I show that these functions g are also in X? And how can I show that every continuous non-differentiable function is in $\overline{D}/\dot{D}$? Can someone help me?

REPLY [5 votes]: Here is a direct way of showing density.
Pick $f \in C[0,1]$. Without loss of generality we can extend $f$ 
to $[-1,2]$
by defining $f(t) = f(0)$ for $t < 0$ and $f(t) = f(1)$ for $t >$1.
Let $f_n(t) = n \int_{t-{1 \over n}}^{t+{1 \over n}} f(\tau) d \tau$.
Then $f_n$ is differentiable with derivative $f_n'(t) = {f(t+{1 \over n})-f(t-{1 \over n}) \over {1 \over n}}$ and it is clear that
$f_n'$ is continuous.
Suppose $\epsilon>0$, then since $f$ is uniformly continuous there is some $\delta>0$ such that if $|x-y| < \delta$ then
$|f(x)-f(y)| < \epsilon$. Now suppose ${2 \over n} < \delta$, then
$|f(t)-f_n(t)| \le n \int_{t-{1 \over n}}^{t+{1 \over n}} |f(t)-f(\tau)| d \tau \le n \int_{t-{1 \over n}}^{t+{1 \over n}} \epsilon d \tau = \epsilon$, and so
$\|f-f_n\|_\infty  \le \epsilon$.