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I am attempting to fill in a proof that $\mathbb R^n$ is a smooth premanifold and its smooth functions are what one would expect: the infinitely differentiable functions from $\mathbb R^n$ to $\mathbb R$. I have the following.

Pick a basis $\mathbf e_1, \dotsc, \mathbf e_n$ for $\mathbb R^n$ and let $x^1, \dotsc, x^n$ be the coordinate functions with respect to this basis. It is clear to me that $x^1, \dotsc, x^n$ are infinitely differentiable, so I will gladly include them in $\mathfrak S(\mathbb R^n)$. Furthermore, it is clear to me that every infinitely differentiable function depends smoothly on $x^1, \dotsc, x^n$, so I now have $\mathcal C^\infty(\mathbb R^n) \subseteq \mathfrak S(\mathbb R^n)$.

My trouble is in showing that other functions don't sneak in. Indeed, couldn't I use the second condition for $\mathfrak S$ to add in a particularly nasty function? All would need to do is modify it to coincide with an infinitely differentiable function on some neighborhood, but leave everything else untouched.

Edit: Correcting the second condition for $\mathfrak S$ removes the ability to add arbitrary functions to $\mathfrak S$, as functions must now coincide near every point of $M$. However, I still have the problem of showing that every member of $\mathfrak S(\mathbb R^n)$ is infinitely differentiable.

Edit 2: Corrected the second condition once again. Furthermore, I believe I have discovered the source of my problem. See my answer below.

My definition of a smooth premanifold is a Hausdorff space $M$ for which the set of all functions $f:M \to \mathbb R$ admits a nonempty subset $\mathfrak S(M)$ of so-called smooth functions satisfying the following two conditions.

  1. If $ f^1, \dotsc, f^r \in \mathfrak S $ and $ f $ depends smoothly on $ f^1, \dotsc, f^r $, then $ f \in \mathfrak S $.
  2. If $g: M \to \mathbb R$ and for each $p \in M$ there exists a function $f \in \mathfrak S(M)$ (possibly depending on $p$) such that $f$ and $g$ coincide near $p$, then $g \in \mathfrak S(M)$.

For thoroughness, two functions coincide near a point if they agree on some neighborhood of that point and a function $f$ depends smoothly on functions $f^1, \dotsc, f^r$ if there exists an everywhere (Edit: this should be infinitely) differentiable function $u: \mathbb R^r \to \mathbb R$ such that $$f(p) = u(f^1(p), \dotsc, f^r(p))$$ for all $p \in M$.

Benjamin
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  • The way you have stated 2. you could indeed add any function and it would have to be smooth. Maybe the source material you're reading has some restrictions of what functions you consider? Or perhaps they only say that $g$ is smooth in that neighbourhood where it coincides with a smooth function? Unless such apply I don't see how $\mathbb{R}^n$ is a smooth premanifold with the usual topology and without considering functions up to certain sets without accepting random functions as smooth which would make ALL functions smooth. – Mr.P Aug 17 '15 at 07:46
  • I think $\mathbb{R}^n$ would be a smooth premanifold with the functions "we would expect to be smooth" with the Zariski topology and if we consider functions up to sets with zero measure. – Mr.P Aug 17 '15 at 07:50
  • @Mr.P I doubled checked condition 2. and I have misstated it. Going to edit the definition now. – Benjamin Aug 17 '15 at 18:44
  • It seems like the new version of the second condition is trivial. If two functions coincide near every point, don't they in particular agree at every point then, hence everywhere? – Ben Aug 17 '15 at 19:21
  • @Ben Yes, yes it is. I've fixed it (once again) so as to be nontrivial. – Benjamin Aug 17 '15 at 20:08
  • In addition, I've discovered that there was a typo in my text in the definition of smooth dependence (see edit and my answer below). I'll leave my answer unchecked for a little while, in case someone else wishes to elaborate on the proof for future viewers. – Benjamin Aug 17 '15 at 20:10

1 Answers1

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Unfortunately, there appears to have been a typo in my text (M. M. Postnikov's The Variational Theory of Geodesics for those curious). Postnikov's definition of a function depending smoothly on some other functions requires the existence of an everywhere differentiable function $u: \mathbb R^r \to \mathbb R$. However, later on infinitely differentiable appears where one would have expected everywhere, if one was following Postnikov's stated definition.

I'm confident that an infinitely differentiable function $u$ was required by the definition. I probably should have been suspicious that so-called smooth dependence only required a $\mathcal C^1$ function.

Either way, since $u$ must be infinitely differentiable to begin with, it is now clear to me that $\mathfrak S(\mathbb R^n)$ cannot contain functions that are not infinitely differentiable.

Xander Henderson
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Benjamin
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  • Just note that authors sometimes say things like "from now on by a smooth function we will mean $C^1$" or $C^\infty$ or "by differentiable we will mean 'sufficiently' differentiable". Those may sometimes be the cause for confusion. – Mr.P Aug 18 '15 at 08:15