Is smooth an open property?

Question

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 $p$

Answer 1

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 $m<n$. 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 $0$.

Answer 2

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.

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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