- #26

- 601

- 7

Let [itex]\gamma :[0,1]\rightarrow\mathbb{R}^{n}[/itex] be a curve in [itex]\mathbb{R}^{n}[/itex] such that [itex]\gamma (0)=a\in\mathbb{R}^{n}[/itex] and [itex]\gamma (1)=x\in\mathbb{R}^{n}[/itex], hence a

*fixed*value of [itex]t\in [0,1][/itex] identifies a point in [itex]\mathbb{R}^{n}[/itex] by [itex]\gamma (t)[/itex]. The curve [itex]\gamma[/itex] is such that it passes through the fixed point [itex]a\in\mathbb{R}^{n}[/itex] and some arbitrary point [itex]x\in\mathbb{R}^{n}[/itex]. The coordinate representation of the curve is given by $$\gamma (t)=\left(x^{1}(t),\ldots,x^{n}(t)\right)=\gamma (0)+t\left(\gamma (1)-\gamma (0)\right)=a+t\left(x-a\right)$$ where [itex]x^{\mu}:[0,1]\rightarrow\mathbb{R}[/itex] are the coordinate functions that specify the coordinates of the curve [itex]\gamma[/itex] in [itex]\mathbb{R}^{n}[/itex] for each value of [itex]t\in [0,1][/itex]. We thus define the coordinate functions [itex]x^{\mu}[/itex] such that $$x^{\mu}(t)=a^{\mu}+t\left(x^{\mu}-a^{\mu}\right)$$ where [itex]a^{\mu}, x^{\mu}\in\mathbb{R}[/itex].

Now consider the function [itex]h:[0,1]\rightarrow\mathbb{R}[/itex], defined such that $$h(t)=(F\circ\gamma)(t)=F(\gamma(t))=F(a+t(x-a))$$ where [itex]F:\mathbb{R}^{n}\rightarrow\mathbb{R}[/itex] is [itex]C^{\infty}[/itex]. It follows that $$\frac{dh}{dt}=h'(t)=\frac{d}{dt}\left[(F\circ\gamma)(t)\right]=\frac{d}{dt}\left[F\left((x^{1}(t),\ldots,x^{n}(t))\right)\right]\\ \qquad=\sum_{\mu =1}^{n}\frac{\partial F}{\partial x^{\mu}}\frac{dx^{\mu}}{dt}=\sum_{\mu =1}^{n}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]\left(x^{\mu}-a^{\mu}\right).$$ Observe that $$h(1)-h(0)=F(x)-F(a)=\int_{0}^{1}h'(t)dt=\int_{0}^{1}\sum_{\mu =1}^{n}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]\left(x^{\mu}-a^{\mu}\right)dt =\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)\int_{0}^{1}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]dt \\ =\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)H_{\mu}(x)$$ where [itex]H_{\mu}(x)= \int_{0}^{1}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]dt[/itex].

Note that [itex] F[/itex] is [itex]C^{\infty}[/itex] and so it follows that [itex] H_{\mu}[/itex] are also [itex]C^{\infty}[/itex].

Finally, taking derivatives of both sides with respect to [itex]x^{\mu}[/itex] we have that $$\frac{\partial F}{\partial x^{\mu}}=H_{\mu}(x)+\left(x^{\nu}-a^{\nu}\right)\frac{\partial H_{\nu}}{\partial x^{\mu}}$$ and so evaluating this at [itex] x=a[/itex] gives $$\frac{\partial F}{\partial x^{\mu}}\bigg\vert_{x=a}=H_{\mu}(a)$$ as claimed.