# Multivariable fundamental calculus theorem in Wald

• I
• sphyrch
In summary: F:\mathbb{R}^n\to\mathbb{R}## is a differentiable function, then$$F(x)=F(a)+\sum_{i=1}^n(x^i-a^i)H_i(x)$$where ##H_i(a)=\frac{\partial F}{\partial x^i}\bigg|_{x=a}##.
sphyrch
i want to prove that if ##F:\mathbb{R}^n\to\mathbb{R}## is a differentiable function, then
$$F(x)=F(a)+\sum_{i=1}^n(x^i-a^i)H_i(x)$$
where ##H_i(a)=\frac{\partial F}{\partial x^i}\bigg|_{x=a}##. the hint is that with the 1-dimensional case, convert the integral into one with limits from ##0## to ##1## and then we'll get the 1-dimensional version of what we're trying to prove. then we have to extend it to ##n##-dimensional case. my try is like this -
$$F(x)=F(a)+\int_a^xF'(s)ds$$
if I substitute ##s=(x-a)t+a##, then the above becomes
$$F(x)=F(a)+(x-a)\int_0^1F'((x-a)t+a)dt$$
so the rhs integral should be my ##H(x)## so that ##H(a)=\frac{dF}{dx}\bigg|_{x=a}=F'(a)##

but first problem: if i evaluate the integral, i get ##H(x)=\frac{F((x-a)t+a)}{x-a}\big|_{t=1}-\frac{F((x-a)t+a)}{x-a}\big|_{t=0}=\frac{F(x)-F(a)}{x-a}## but i don't see how ##H(a)=\frac{dF}{dx}\bigg|_{x=a}=F'(a)##

second problem is, how should I extend to the ##n##-dimensional case? the most I can think of is that ##n##-dimensional ##F## will have several component functions ##F_1,\ldots,F_n## - to each of which we can apply the 1-D result, but how does that get us to the final result? Please help

Last edited:
$$x_1,x_2\in\mathbb{R}^m,\quad\int_0^1\frac{d}{ds}f(sx_1+(1-s)x_2)ds=f(x_1)-f(x_2);$$
$$\int_0^1\frac{d}{ds}f(sx_1+(1-s)x_2)ds=\int_0^1\frac{\partial f}{\partial x}(sx_1+(1-s)x_2)ds(x_1-x_2);$$
$$H_i=\int_0^1\frac{\partial f}{\partial x^i}(sx_1+(1-s)x_2)ds$$
ok?

Last edited:
vanhees71

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