Mean value theorem
Mean value theorem in several variables

The mean value theorem generalizes to real functions of multiple variables. The trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.

Let G {\displaystyle G} be an open subset of R n {\displaystyle \mathbb {R} ^{n}}, and let f: G → R {\displaystyle f:G\to \mathbb {R} } be a differentiable function. Fix points x, y ∈ G {\displaystyle x,y\in G} such that the line segment between x, y {\displaystyle x,y} lies in G {\displaystyle G}, and define g (t) = f ((1 − t) x + t y) {\displaystyle g(t)=f{\big (}(1-t)x+ty{\big)}}. Since g {\displaystyle g} is a differentiable function in one variable, the mean value theorem gives:

for some c {\displaystyle c} between 0 and 1. But since g (1) = f (y) {\displaystyle g(1)=f(y)} and g (0) = f (x) {\displaystyle g(0)=f(x)}, computing g ′ (c) {\displaystyle g'(c)} explicitly we have:

where ∇ {\displaystyle \nabla } denotes a gradient and ⋅ {\displaystyle \cdot } a dot product. This is an exact analog of the theorem in one variable (in the case n = 1 {\displaystyle n=1} this is the theorem in one variable). By the Cauchy–Schwarz inequality, the equation gives the estimate:

In particular, when the partial derivatives of f {\displaystyle f} are bounded, f {\displaystyle f} is Lipschitz continuous (and therefore uniformly continuous).

As an application of the above, we prove that f {\displaystyle f} is constant if the open subset G {\displaystyle G} is connected and every partial derivative of f {\displaystyle f} is 0. Pick some point x 0 ∈ G {\displaystyle x_{0}\in G}, and let g (x) = f (x) − f (x 0) {\displaystyle g(x)=f(x)-f(x_{0})}. We want to show g (x) = 0 {\displaystyle g(x)=0} for every x ∈ G {\displaystyle x\in G}. For that, let E = { x ∈ G: g (x) = 0 } {\displaystyle E=\{x\in G:g(x)=0\}}. Then E is closed and nonempty. It is open too: for every x ∈ E {\displaystyle x\in E},

for every y {\displaystyle y} in some neighborhood of x {\displaystyle x}. (Here, it is crucial that x {\displaystyle x} and y {\displaystyle y} are sufficiently close to each other.) Since G {\displaystyle G} is connected, we conclude E = G {\displaystyle E=G}.

The above arguments are made in a coordinate-free manner; hence, they generalize to the case when G {\displaystyle G} is a subset of a Banach space.
combination:two_responses, startend:end_checker, punctuation:no_comma