Let $X : C^{\infty}(\mathbb{R}^n) \rightarrow \mathbb{R}$ be a derivation, so i.e. linear and satisfying the Leibniz Rule $$X(fg)=X(f) \cdot g(a)+X(g) \cdot f(a)$$ for some fixed $a \in \mathbb{R}^n.$ Then I would like to see that $X = \sum_{i=1}^{n} a_i \frac{\partial}{\partial x_i}|_a.$ So I want to see that $X$ is essentially a directional derivative.
Asked
Active
Viewed 532 times
3
-
Here, ℂ∞(ℝn) is all continuous derivatives of all orders?? – Apr 23 '15 at 00:05
-
@J.G yes, I adjusted the notation a little bit. – RealAnalysis Apr 23 '15 at 00:09
-
I'm not sure of your question here. You want us to solve the function with X=other operation? – Apr 23 '15 at 00:16
-
@J.G. I want you to show this or give me a suitable hint :-) – RealAnalysis Apr 23 '15 at 00:23
-
Expand in Taylor series and use the fundamental theorem of calculus. You just need this. – user40276 Apr 23 '15 at 00:25
-
@user40276 not every smooth function has a taylor series. – RealAnalysis Apr 23 '15 at 00:27
-
Every smooth function have a Taylor series with an integral remainder. You have to use this. – user40276 Apr 23 '15 at 01:01
-
@RealAnalysis: Hi RealAnalysis. Is there something wrong with my solution below? Just wanted to ask as you haven’t mentioned anything about it. – Berrick Caleb Fillmore Apr 23 '15 at 14:19
1 Answers
3
As user40276 has mentioned, using Taylor’s Theorem, we can write $$ \forall \mathbf{x} \in \mathbb{R}^{n}: \quad f(\mathbf{x}) = f(\mathbf{a}) + \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} \Bigg|_{\mathbf{x} = \mathbf{a}} \cdot (x_{i} - a_{i}) + R(\mathbf{x}), $$ where $ R: \mathbb{R}^{n} \to \mathbb{R} $ is a linear combination of products of smooth functions that vanish at $ \mathbf{a} $.
Hence, if $ \Theta_{i} $ denotes the map $ \mathbf{x} \mapsto x_{i} - a_{i} $, then by the linearity of $ X $, the Leibniz Rule and also the fact that $ X $ sends constant functions to $ 0 $, we obtain $$ \forall f \in {C^{\infty}}(\mathbb{R}^{n}): \quad X(f) = \sum_{i = 1}^{n} X(\Theta_{i}) \cdot \frac{\partial f}{\partial x_{i}} \Bigg|_{\mathbf{x} = \mathbf{a}}. $$
Berrick Caleb Fillmore
- 5,269