# Are all derivations of real-valued functions derivatives?

To me it was obvious that all derivatives are derivations, since one can show that they are linear and satisfy the product rule. However, I was very surprised when I learned that every derivation of $C^{\infty}$ functions of a (smooth) manifold is essentially a directional derivative. The definition of a derivation is very simple and consists of two purely algebraic properties, while the definition of directional derivative should seemingly be inseparable from purely analytic notions like limits, Caucy/metric completeness, and the least upper bound property (although these notions may actually be purely topological, see here, although in any case they are not algebraic).

However, the proof (see my community-wiki "answer" below) relies in a seemingly essential way on Taylor's theorem and its generalizations, and thus on the twice differentiability of the functions.

This would seem to explain why calculus is taught in terms of derivatives and not derivations, in spite of the latter's greater simplicity, because to show that they are equivalent requires Taylor's theorem, which to understand requires a prior understanding of what differentiation is, and in any case is usually taught first in the second semester.

Question: When we require only first differentiability and not second differentiability or continuous differentiability, is it still the case that all derivations are differential operators?

Are there no derivations defined on the space of continuous real-valued functions?

I.e., in every context where both derivatives and differentiation are defined, are derivations really exactly the same thing as derivatives? Or only when derivations are restricted to those functions for which Taylor's theorem holds?

I have heard of there existing derivations for fields that are not even metrically complete, such as the finite fields $\mathbb{F}_p$, and since the definition is really so much simpler than that of the derivative, I would be really surprised if they coincided exactly whenever both existed.

Related questions: (1)(2)(3)

Also I am not sure how to tag this question so please feel free to fix the tags.

• It's important to note that derivations only turn out to be directional derivatives in the space of smooth (i.e. $C^\infty$) functions. In general it is false, see ams.org/journals/bull/1973-79-04/S0002-9904-1973-13293-8/… – Pedro Nov 5 '16 at 15:20
• @Pedro This is a very nice and informative result which answers my question more or less entirely -- I thank you for referring it to me, because I did not know it earlier. Also, more as a comment to myself, the definition of tangent space used in that paper resembles the Zariski definition used sometimes in algebraic geometry, which is interesting to me. – Chill2Macht Nov 5 '16 at 16:31

Let $\mathcal{C}^{\infty}(\mathbb{R}^n)$ denote the space of all smooth (scalar-valued) functions $\mathbb{R}^n \to \mathbb{R}$. Note that $\mathcal{C}^{\infty}$ satisfies all of the properties of an $\mathbb{R}-$module.
A derivation at $a \in \mathbb{R}^n$ is a map $w_a: \mathcal{C}^{\infty} \to \mathbb{R}$ satisfying the following two properties: $$w_a(\lambda_1f_1+\lambda_2f_2)=\lambda_1w_a(f_1)+\lambda_2w_a(f_2) \quad \forall\ \lambda_1,\lambda_2 \in \mathbb{R}, \forall\ f_1,f_2 \in \mathcal{C}^{\infty} \\ w_a(fg)=f(a)w_a(g)+w_a(f)g(a)$$ If $\varepsilon_a$ denotes evaluation at $a$, i.e. $f \mapsto f(a)$, then $w: \mathcal{C}^{\infty} \to \mathcal{C}^{\infty}$ is a derivation if $\varepsilon_a \circ w$ is a derivation at $a$ for all $a \in \mathbb{R}^n$.
Let $w_a$ be an arbitrary derivation at $a$, and let $\pi_i$ denote the $i$th coordinate projection function $\mathbb{R}^n \to \mathbb{R}: (x_1, \dots, x_n) \mapsto x_i$. Denote $v^i:=w_a(\pi_i)$, then if $e_1, \dots, e_n$ are the standard unit vectors for $\mathbb{R}^n$, we define $v:= \sum_i v^i e_i$. We will show that $w_a$ is the directional derivative of $v$ evaluated at $a$, i.e. $$w_a = \varepsilon_a \circ D_v = D_v \rvert_a.$$ From this it will follow that every derivation is a directional derivative of some vector in $\mathbb{R}^n$.
Let $f \in \mathcal{C}^{\infty}$. Using Taylor's theorem (see below) we have that $\forall x \in \mathbb{R}^n$, $$\begin{array}{rcl} f(x) & = & f(a) + \displaystyle\sum_{i=1}^n \frac{\partial f}{\partial x^i}(a)(\pi_i(x)-\pi_i(a)) \\& & + \displaystyle\sum_{i,j=1}^n (\pi_i(x) -\pi_i(a))(\pi_j(x) - \pi_j(a)) \int\limits_0^1 (1-t)\frac{\partial^2 f}{\partial x^i \partial x^j}(a+t(x-a))\mathrm{d}t. \end{array}$$ Defining $g_{ij}(x)=\displaystyle\int\limits_0^1 (1-t)\frac{\partial^2 f}{\partial x^i \partial x^j}(a+t(x-a))\mathrm{d}t$, we evaluate $w_a(f)$ as follows: $$w_a(f) = w_a\left( f(a) + \sum\limits_{i=1}^n \frac{\partial f}{\partial x^i}(a) (\pi_i - \pi_i(a)) + \sum\limits_{i,j=1}^n (\pi_i - \pi_i(a))(\pi_j-\pi_j(a))g_{ij} \right) \\ = w_a(f(a))+\sum\limits_{i=1}^n \frac{\partial f}{\partial x^i}(a)w_a\left( (\pi_i - \pi_i(a))\right)+\sum\limits_{i,j=1}^n w_a((\pi_i - \pi_i(a))(\pi_j - \pi_j(a))g_{ij} ) \\ = f(a)w_a(1) + \sum\limits_{i=1}^n \frac{\partial f}{\partial x^i}(a)\left[w_a(\pi_i) - \pi_i(a)w_a(1) \right] + \sum\limits_{i,j=1}^n \left[ (\pi_i - \pi_i(a))\rvert_a \cdot w_a((\pi_j -\pi_J(a))g_{ij})+w_a(\pi_i - \pi_i(a)) \cdot ((\pi_j-\pi_j(a))g_{ij})\rvert_a \right] \\ = f(a)\cdot0 + \sum\limits_{i=1}^n \frac{\partial f }{\partial x^i}(a)[w_a(\pi_i)-\pi_i(a)\cdot 0] + \sum\limits_{i,j=1}^n \left[ 0 \cdot w_a((\pi_j-\pi_j(a))g_{ij}) + w_a(\pi_i - \pi_i(a))\cdot 0 \right] \\ = 0 + \sum_{i=1}^n \frac{\partial f}{\partial x^i}(a)[w_a(\pi_i)-0] + \sum_{i,j=1}^n [0+0] = \sum_{i=1}^n \frac{\partial f}{\partial x^i}(a) v^i = D_v (f)\rvert_a.$$ Hence $w_a = D_v \rvert_a$, as was claimed.