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.
- If $ f^1, \dotsc, f^r \in \mathfrak S $ and $ f $ depends smoothly on $ f^1, \dotsc, f^r $, then $ f \in \mathfrak S $.
- 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 $p \in M$.
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.