Is $(\mathcal{C}^1([a,b])$ dense is $(L^{\infty},\lVert\cdot\rVert_{L^{\infty}})$

functional-analysislp-spacessolution-verification

We know $\mathcal{C}^0([a,b])$ is a closed subset of $(L^{\infty},\lvert\cdot\rVert_{L^{\infty}})$, and moreover for any continuous function $\lVert f \rVert_{\infty}=\lVert f \rVert_{L^{\infty}}$.

My thoughts: $(\mathcal{C}^1([a,b]),\lVert\cdot\rVert_{L^{\infty}})$ is open, but not dense (and moreover its closure, under $\lVert \rVert_{L^{\infty}}$, coincide with $C^0([a,b])$).

Proof:

  1. $\mathcal{C}^1([a,b])$ is open: since for any continuous function the $L^\infty$-norm and the supremum norm coincide, this space is not closed as sequences like $(|x|^{1+\frac{1}{h}})_h$ (ok, in $[0,1]$, but you can modify as you wisk to make them fit in any $[a,b]$) converge to a a function whose derivative is discontinuous;
  2. $\mathcal{C}^1([a,b])$ is not dense: surely its closure contains $\mathcal{C}^0([a,b])$, thanks to Stone-Weierstrass, but since $\mathcal{C}^0([a,b])$ is not dense in $(L^{\infty},\lVert\cdot\rVert_{L^{\infty}}$), and we're working with one of its subsets, we may conclude;
  3. $\overline{\mathcal{C}^1([a,b])}^{\lVert\cdot\rVert_{L^{\infty}}}=\mathcal{C}^0([a,b])$: again, it's obvious the closure contains this set, but since $\mathcal{C}^0([a,b])$ is closed in $(L^{\infty},\lVert\cdot\rVert_{L^{\infty}}$) (as we've said at the beginning), we may conclude.

Edit: While writing it turns out I was more or less able to answer the question: I thought this was a question full of question marks, so I kindly ask you to review it and let me know if I'm missing something very important, or the proof are fine. This was just a question popped out of my mind, so I've not been particularly rigorous.

Thanks to anyone who will spend some minutes helping me.

Best Answer

$2$ and $3$ are correct, $1$ is mostly correct, $C^1([a,b])$ is definitely not closed in the $\|\cdot\|_\infty$ norm, but it is not open either.

Not only it is not open, it is actually a set with empty interior: For every $f\in C^1([a,b])$ and every $\varepsilon>0$ you can easily find a discountinuous function $g\in L^\infty([a,b])$ with $\|f-g\|_\infty<\varepsilon$, so $C^1([a,b])$ contains no balls. In fact the same reasoning applies to $C^0([a,b])$, which is the closure of $C^1([a,b])$ by $3$, so $C^1([a,b])$ is even nowhere dense in $L^\infty([a,b])$.

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