# $C^1[a,b]$ is closed in $C[a,b]$

So, I have the following question:

Let $C[a,b]$ denote the space of continuous real-valued functions on $[a,b]$ with the sup-metric. Let $C^1[a,b]$ denote the space of continuously differentiable functions on $[a,b]$. Is $C^1[a,b]$ a closed subset of $C[a,b]$?

So, what I've done so far. The sup metric is $d_{\infty}(f,g) = \sup \left\{|f(x) - g(x)| : a \leq x \leq b \right\}$ for $f,g\in C[a,b]$

Clearly $C^1[a,b] \subset C[a,b]$

$C^1[a,b]$ is a closed subset of $C[a,b]$ if $C[a,b]\backslash C^1[a,b]$ is open in $C[a,b]$ - i.e.

$\forall h \in C[a,b]\backslash C^1[a,b]$, $\exists \epsilon > 0$ such that $\left\{ f \in C[a,b] : d_{\infty}(f,h) < \epsilon \right\} \subseteq C[a,b]\backslash C^1[a,b]$

Which seems to be saying that, essentially, for any non-differentiable continuous function $h$ we can find an $\epsilon > 0$ such that when the supremum of the distance between $h$ and all other continuous functions on $[a,b]$ is less than $\epsilon$, then they will also not be differentiable.

This is where I get stuck, not least because I'm not convinced whether that is true or not. Clearly, it's easily possible to create further non-differentiable functions by, for example, moving one point of $h$ less than $\epsilon$. My instinct is that you're not going to be able to turn a non-differentiable function into a differentiable one with an arbitrarily small $\epsilon$, so I think such an $\epsilon$ exists - but this obviously isn't formal (or even likely to be correct)

Thanks.

No -- on the contrary, $C^1[a,b]$ is dense in $C[a,b]$, by the Weierstrass approximation theorem. But that's too big a gun to bring out to show that it is simply not closed.

Instead, to follow your ansatz, Karolis' suggestion is to consider $h(x)=|x|$. That's the archetypical example of a function that's continuous but not differentiable, if $0$ is an interior point of $[a,b]$ (which we can clearly assume without loss of generality).

We're then looking for an $\epsilon$ such that no $f$ that differs from $h$ by at most $\epsilon$ can be differentiable. If we fail in that -- in other words, if for every $\epsilon$ there is a differentiable $f$ close to $h$ -- then we have proved that $C^1[a,b]$ is not closed.

Now consider $f(x)=\sqrt{x^2+\delta}$ where $\delta$ is a small positive constant that possibly depends on $\epsilon$. Is this always differentiable? Can we made it stay within a distance of $\epsilon$ from $|x|$ by choosing $\delta$ small enough?

• On an interval $[a,b]$ you can make it stay within a distance of $\epsilon$ from $|x|$, but not if you considering the cases where $x$ goes to $\pm \infty$. – Noble. Oct 18 '13 at 18:20
• @Noble: Fortunately the relevant metric here considers only differences in function values on $[a,b]$. But actually this argument does work for $C^1(\mathbb R)$ and $C(\mathbb R)$ too, because $\sqrt{x^2+\delta} - |x| \approx \frac{\delta}{2 |x|}$ for large $x$. – Henning Makholm Oct 18 '13 at 18:35
• Right, and so because this $f(x)$ is differentiable, we've found a counterexample essentially, so it isn't closed? – Noble. Oct 18 '13 at 19:04
• @Noble: Correct. – Henning Makholm Oct 18 '13 at 19:17

Try building a sequence of differentiable functions, converging to $f(x) = |x|$ (say $[a, b] = [-1, 1]$).

• Sorry, I don't think I'm fully understanding the suggestion. – Noble. Oct 18 '13 at 16:54
• A set is closed iff every convergent sequence of points in that set converges to a point in that set. $|x|$ is a function differentiable except for a single point. If you, say, replace that corner with a quarter of a circle, it becomes differentiable. If the radius of that circle is tiny, then by sup metric, the functions are very close. Thus you have a sequence. – Karolis Juodelė Oct 19 '13 at 7:10