The interior of $C^1([0,1])$ is empty

Question

The interior of $C^1([0,1])$ w.r.t. $\Vert \cdot \Vert_{\infty}$ is empty

My intuition is that we have to use the fact that $C([0,1])$ is dense in $C^1([0,1])$. But I don't really know how to formulate a solution formally.

My solution: Assume it is not empty. Then it exists $f\in C^1([0,1])$ s.t. it exists $U$ containing $f$ (a neighborhood of $f$) with $U\in C^1([0,1])$. On the other hand since $C([0,1])$ is dense in $C^1([0,1])$, for this fixed $f$ is exists a sequence $f_n\in C([0,1])\backslash C^1([0,1])$ s.t. $\Vert f_n-f\Vert_{\infty}\leq \varepsilon$ but then for $n$ big enough $f_n\in U\in C^1([0,1])$ and we have a contradiction!

Is my solution ok? Is it a more beautiful way? Thank you for your help.

Ah sorry: is empty with respect to the $\Vert \Vert_{\infty}$-norm i.e. the sup-norm. thank you @Dark — MorganeMaPh, Aug 01 '16 at 13:54
I don't see how you can choose your $f_n$. And what do you mean a larger set $C([0,1])$ is dense in a smaller set $C^1([0,1])$? — GEdgar, Aug 01 '16 at 14:11

score 0 · Accepted Answer · answered Aug 01 '16 at 16:21

By contradiction, let $f \in U$ where $U$ is an open subset of $C^1([0,1])$.

There exists $\epsilon>0$ such that for all $g$, $||f-g|| \leq \epsilon \implies g \in C^1([0,1])$

Let $h \in C([0,1]) \setminus C^1([0,1])$ and $g=f+\epsilon \frac{h}{||h||}$.

Then $||g-f||=\epsilon$ so $g \in C^1([0,1])$.

Hence $h=\frac{||h||}{\epsilon}(g-f)$ so $h \in C^1([0,1])$. Contradiction.

The interior of $C^1([0,1])$ is empty

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