vanishing of gradient in domain

Theorem.  If the function $f$ is defined in a domain (http://planetmath.org/Domain2) $D$ of ${ℝ}^n$ and all the partial derivatives of a $f$ vanish identically in $D$, i.e.

$$\nabla f\equiv \overrightarrow{0}\mathit{\hspace{1em}}\text{in}D,$$

then the function has a constant value in the whole domain.

Proof.  For the sake of simpler notations, think that  $n=3$; thus we have

$f_x^{\prime }(x,y,z)=f_y^{\prime }(x,y,z)=f_z^{\prime }(x,y,z)=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{for all}(x,y,z)\in D.$ (1)

Make the antithesis that there are the points  $P_0=(x_0,y_0,z_0)$  and  $P_1=(x_1,y_1,z_1)$  of $D$ such that  $f(x_0,y_0,z_0)\ne f(x_1,y_1,z_1)$.  Since $D$ is connected, one can form the broken line $P_0{Q}_1{Q}_2\mathrm{\dots }{Q}_k{P}_1$ contained in $D$.  When one now goes along this broken line from $P_0$ to $P_1$, one mets the first corner where the value of $f$ does not equal $f(x_0,y_0,z_0)$.  Thus $D$ contains a line segment, the end points of which give unequal values to $f$.  When necessary, we change the notations such that this line segment is $P_0{P}_1$.  Now, $f_x^{\prime },f_y^{\prime },f_z^{\prime }$ are continuous in $D$ because they vanish.  The mean-value theorem for several variables guarantees an interior point  $(a,b,c)$  of the segment such that

$$0\ne f(x_1,y_1,z_1)-f(x_0,y_0,z_0)=f_x^{\prime }(a,b,c)(x_1-x_0)+f_y^{\prime }(a,b,c)(y_1-y_0)+f_z^{\prime }(a,b,c)(z_1-z_0).$$

But by (1), the last sum must vanish.  This contradictory result shows that the antithesis is wrong, which settles the proof.

Title	vanishing of gradient in domain
Canonical name	VanishingOfGradientInDomain
Date of creation	2013-03-22 19:11:58
Last modified on	2013-03-22 19:11:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26B12
Synonym	partial derivatives vanish
Related topic	FundamentalTheoremOfIntegralCalculus
Related topic	ExtremumPointsOfFunctionOfSeveralVariables