# Is smooth an open property?

Surely this is in some textbook somewhere, but I wasn't able to find it.

Open property: If property true at $$p$$, then true at open subset containing $$p$$.

1. For $$f: \mathbb R \to \mathbb R$$, continuous, real differentiable and real twice differentiable aren't open properties.

2. Holomorphic $$f: \mathbb C \to \mathbb C$$, complex-analytic $$f: \mathbb C \to \mathbb C$$ and real-analytic $$f: \mathbb R \to \mathbb R$$ are (and real-holomorphic $$f: \mathbb R \to \mathbb R$$ too defined as real differentiable in an open interval containing $$p$$) open properties.

Question: Is smooth an open property, i.e. if a function $$f: \mathbb R \to \mathbb R$$ is smooth at a point $$p$$, then is it smooth on an open subset containing $$p$$?

• Here: smooth at $$p$$ is defined at $$C^k$$ at $$p$$ for all $$k \ge 0$$. $$C^k$$ at $$p$$ in turn is defined as all derivatives of orders $$0 \le j \le k$$ exist and are continuous at $$p$$
• Definition of smooth? Oct 13, 2021 at 21:26
• @MarkSaving thanks. if you're asking me to define: then edited. if you're asking if this follows simply by definition of smooth: then please see answer
– BCLC
Oct 13, 2021 at 21:55

No! The problem with your proof is that $$U$$ depends on $$n$$, and could thus get smaller with rising $$n$$.

Consider this counter example: Let $$k_n$$ be a function so that $$k_n$$ is $$n-1$$ times continously differentiable on $$\mathbb R$$ and $$n$$ times c.d. only on $$(-1/n,1/n)$$. Also we want $$k_n^{(m)}<1/2^{n-m}$$ for $$0\leq m\leq n$$ and $$k^{(m)}=0$$ outside of $$[-1/n,1/n]$$ for $$m. For example we can take some function like $$\chi_{[-1/n,1/n]}(x)(1/n^2-x^2)^n$$, scale in such a way that the bound to the derivatives hold.

Define $$f_m(x) = \sum_{n=0}^m k_n$$. It is clear to see that $$f_m$$ converges uniformly to some $$f$$ (as $$\sum_{n=m+1}^\infty |k_n| < \sum_{n=m+1}^\infty 1/2^{m+1} \frac{1}{1+1/2} = 1/2^m$$).

Also we get that if $$m>n$$ then $$f_m$$ is $$n$$-times differentiable on $$(-1/n,1/n)$$. Also it is equal to $$f_{m-1}$$ outside of $$[-1/(m+1),1/(m+1)]$$. Thus it is only $$n-1$$ times differentiable in $$\pm 1/n$$. Then it is easy to see that $$f_m^{(n)}$$ converges uniformly against some $$g_n$$ within $$(-1/n,1/n)$$ (for the same reason as to why $$f_m$$ converges).

But this implies that $$f$$ is $$n$$ times differentiable on $$(-1/n,1/n)$$ so that $$f^{(n)}=g_n$$. On the other hand $$f$$ is equal to $$f_n$$ outside of $$[-1/{n+1},1/{n+1}]$$, so it is only $$n-1$$ times differentiable in $$\pm 1/n$$.

But thus we get that $$f$$ is smooth in $$x=0$$, but only $$n-1$$ times differentiable in $$\pm1/n$$ and thus not smooth an any neighborhood of $$0$$.

• thanks Lazy but smooth does imply real-holomorphic (real-holo at p means real differentiable in an open interval containing p)? and in general it does imply both $k$-times differentiable and continuous $k$-times differentiable for any (finite! hehe) $k$?
– BCLC
Oct 13, 2021 at 23:25
• Well, higher order derivatives don’t make a lot of sense if you do not actually have a derivative around $p$.
– Lazy
Oct 13, 2021 at 23:51
• Lazy, btw I asked full question here: 2 basic quick questions about smooth, continuously differentiable, real-holomorphic.
– BCLC
Oct 27, 2021 at 17:25

Edit:

1. This is actually wrong. oh thank God i'm wrong. i thought this was some simple or easy thing i was too dumb to see.

2. Ok so it seems the reason why this is wrong is that what I've done is just get that $$f$$ is continuous $$n$$-times differentiable on $$U_n (\ni p)$$ so $$f$$ is smooth on $$U = \cap_{n=1}^{\infty} U_n (\ni p)$$, but $$U$$ need not be open.

Oh wait I think I figured it out from this: Why twice differentiable at a point implies differentiable in an interval (which I like to call real-holomorphic)

Yes.

Proof:

1. Twice differentiable at $$p$$ means differentiable on an open subset $$U$$ containing $$p$$. (Maybe we can strengthen the conclusion to continuously differentiable but no need to assume this. Just do instead...)
2. (1) extends to that $$(n)$$-times differentiable at a point implies $$(n-1)$$-times differentiable on $$U$$.
3. By(2), thrice differentiable at $$p$$ means twice differentiable on $$U$$. Then we get continuous differentiable on $$U$$.
4. (3) extends to that $$(n)$$-times differentiable at $$p$$ means $$(n-1)$$-times differentiable on $$U$$ and thus continuous $$(n-2)$$-times differentiable on $$U$$.
5. Therefore, smooth at $$p$$ implies $$(n)$$-times differentiable at $$p$$ and thus continuous $$(n-2)$$-times differentiable on $$U$$

QED

• This doesn't work. You've shown that for all $n$, there exists a neighbourhood $U$ of $p$ such that $f$ is $n$-times continuously differentiable on $U$. But you haven't picked a single $U$ such that for all $n$, $f$ is $n$-times continuously differentiable on $U$. This is because the intersection of a countable collection of open sets is not necessarily open. Oct 13, 2021 at 21:58
• @MarkSaving oh thank God i'm wrong. i thought this was some simple or easy thing i was too dumb to see.
– BCLC
Oct 13, 2021 at 22:00