In my experience, people usually use "smooth" to mean "as smooth as I need for the upcoming proofs." Those who want to be more formal might insist on smooth meaning $C^\infty$.
While the operator taking $f$ to its Taylor series at some point in its domain uses information about all the partial derivatives of $f$ at once, I don't think I know an example to settle the following question:
Is there anything I can do simultaneously with infinitely many derivatives of a $C^\infty$, not necessarily analytic function?
I hope that's not phrased too vaguely. Since there are $C^\infty$ functions with zero Taylor series, for instance, something like "sure-you can put all its derivatives into an infinite series!" is probably not relevant. I won't be surprised if answer is "No, $C^\infty$ doesn't mean materially more than $C^k$ for sufficiently large $k$." But I'd be very interested to hear otherwise.
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$\begingroup$I found something interesting! Let $M$ be a $C^k$ manifold, $1\leq k\leq \infty$ and $\mathcal{O}(M)_P$ its local ring of germs of $C^k$ functions at some point $P\in M$. One definition of tangent vector is a derivation on $\mathcal{O}(M)_P$, that is, a linear map $\alpha:\mathcal{O}(M)_P\to \Bbb{R}$ such that $\alpha(fg)=\alpha fg(P)+f(P)\alpha(g)$. Then it becomes necessary to prove that the vector space $A_p$ of derivations of $\mathcal{O}(M)_P$ has the same dimension as $M$.
In case $k=\infty,$ one can do this (following Warner, Foundations of Differentiable Manifold and Lie Groups) by indicating an isomorphism between $A_p$ and $(\mathfrak{m}_P/\mathfrak{m}^2_P)^*$, where $\mathfrak{m}_P$ is the maximal ideal of $\mathcal{O}(M)_P,$ i.e. the space of germs of functions vanishing at $P$. This isomorphism exists whether $k$ is finite or infinite.
But then we have to show $\mathfrak{m}_P/\mathfrak{m}^2_P$ is $n$-dimensional, and for $k$ finite this is false! To prove this in case $k=\infty$ one just obsereThe proof is an extension of the simple calculus exercise that shows every smooth map $f:\Bbb{R}\to\Bbb{R}$ with $f(0)=0$ is divisible by $x$, i.e. the coordinate function $x$ generates the ideal $\mathfrak{m}_0(\Bbb{R})$. One does this by observing that $f=xg$ for $g(x)=\int_0^1 f'(tx)dt$. This is the point where we need infinite $k$: for $k<\infty,$ $g$ need not be a $C^k$ function!
It actually turns out that for $C^k$ manifolds with finite $k$, the space of derivations $A_P$ is infinite-dimensional, and so can't serve as a model of tangent space at all. The relevant issue is exactly that the product of two $C^k$ functions vanishing at $P$ is in $C^{k+1}$ near $P$; this permits derivations which don't vanish on every $C^k$ function with zero partial derivatives.
$\endgroup$ 2 $\begingroup$One very useful fact is that if $G$ is an open subset of $\mathbb{R}^d$, then $\mathcal{C}^\infty_c(G)$ is dense in $\mathcal{L}^p(G)$ if $1\le p < \infty$.
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