Two functions that have the same germ have the same derivation

Question

Is this proof okay? (EDIT: is the claim even true?)

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I claim: $f$ and $g$ have the same germ at $p$ if and only if they have the same derivations at the same point.

($f$ and $g$ has the same germ at $a$ if for $x$ in some neighborhood of $a$ we have $f(x)=g(x)$. A derivation (at $p$) is a map $X\colon C^\infty(M) \to \mathbb R$ satisfying the product rule $X(fg) = f(p)\cdot Xg + g(p)\cdot Xf$. It's established elsewhere that the set of all germs at $p$ and the set of all derivations at $p$ are vector spaces.)

First we show that if two functions have the same derivation, then they must have the same germ. We proceed throughout inside a neighborhood $W$ of $p$.

By linearity, two functions have the same derivation if it is true that $Xp(f-g)=0$. For $\mathtt X$ to be an actual derivation, it must hold that for any other function $k$ defined in $W$ $$ X [(f-g)\cdot k] (w) = k(w) \cdot X (f-g) (w) + (f-g)(w)\cdot X k(w) $$

The derivation of a function that is constant and equal to zero (i.e. $(f-f)$ for any $f$) must be zero everywhere. Therefore, the first term in the right hand side is zero. Moreover, $(f-g)(w) \cdot k(w) = 0$ whenever $(f-g)(w) = 0$, which makes the left hand side zero as well. This leaves $$ (f-g)(w)\cdot X k(w) = 0 $$ For this formula to hold for an arbitrary function $k$, we must have $(f-g)=0$ in $W$. Therefore these two functions have the same germ.

The converse may be shown by a contradiction argument. Let $f, g$ be functions with the same germ near $p$, but different derivations. Then, in a neighborhood of $p$ we have that $X(f) - X(g) = X(f-g) = c \neq 0$. But since these functions have the same germ, in a (possibly smaller) neighborhood of $p$ there must be $(f-g)=0$ and therefore $X(f-g) = 0$. Thus, two functions that have the same germ also have the same derivation near the same point.

Answer 1

No, the claim is not true.

It's not clear exactly what you mean when you say two functions $f$ and $g$ "have the same derivations at the same point." I can think of two possible interpretations for this phrase:

1. For every derivation $X$ at the point $p$, we have $Xf = Xg$. [This seems to be what you mean by the phrase "same derivations at the same point."]
2. For every derivation $X$ at every point in some neighborhood $W$ of $p$, we have $Xf = Xg$. [This might be what you're actually using in your argument, but it's not clear.]

In any case, the statement you're trying to prove is false no matter which of these interpretations you had in mind.

If you use interpretation #1, then the statement is wildly untrue. For example, just take $M$ to be the real line, $p$ to be the origin, and consider the functions $f(x) = 0$ and $g(x) = x^2 +1$. Every derivation at $0$ is some constant multiple of the derivative operator $d/dx$. Because $f'(0)=g'(0)=0$, it follows that $f$ and $g$ satisfy condition 1, but they are not equal anywhere so they don't have the same germ at $0$.

If you use interpretation #2, the possibilities are more limited, but here's a counterexample: Again take $M=\mathbb R$ and $p=0$, but this time let $f(x)=0$ and $g(x)=1$. Now every derivation at every point gives the same result when applied to $f$ and $g$, but $f$ and $g$ are not equal anywhere.

What you can say is that if two functions satisfy condition #2, then their difference is locally constant; that is, on each connected component of $M$, there is a constant $c$ such that $f = g+c$.

I think the problem with your argument is that in both parts of the argument, you seem to be assuming that $f-g$ is identically zero in some neighborhood $W$. This is the correct assumption for the second part of the argument (same germ implies equal derivations), but by assuming this in the first part you've made your argument circular.