Properties of Smooth Vector Field

Calculus Level pending

Any smooth vector field $X:\mathbb R^n\to\mathbb R^n$ yields a map $\tilde X:C^\infty(\mathbb R^n)\to C^\infty(\mathbb R^n),$ defined by $(\tilde Xf)(p)=D_{X(p)}f(p)=\lim_{t\to 0}\frac {f(p+tX(p))-f(p)}t.$ This operation is linear over $\mathbb R$ and satisfies the product rule: $\tilde X(fg)=\tilde X(f)g+f\tilde X(g).$ Conversely, if a map $D:C^\infty(\mathbb R^n)\to C^\infty(\mathbb R^n)$ is linear over $\mathbb R$ and satisfies the following product rule: $D(fg)=D(f)g+fD(g),$ must $D$ be $\tilde X$ for certain smooth vector field $X$ ?

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Otto Bretscher
Nov 16, 2018

It is a well-known fact from vector calculus that $\omega(f)=\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}\omega(x_i)$ for any linear operator $\omega$ on $C^{\infty}(\mathbb{R}^n)$ that satisfies the product rule. But this means that $\omega(f)=\nabla f \cdot X=\tilde X(f)$ , the derivative in the direction of $X$ , where $X$ is the vector field whose component functions are the $\omega(x_i)$ for $i=1,..n$ .

Properties of Smooth Vector Field

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