vanishing of gradient in domain

Theorem.  If the function f is defined in a domain ( D of n and all the partial derivativesMathworldPlanetmath of a f vanish identically in D, i.e.


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

fx(x,y,z)=fy(x,y,z)=fz(x,y,z)= 0for all(x,y,z)D. (1)

Make the antithesis that there are the points  P0=(x0,y0,z0)  and  P1=(x1,y1,z1)  of D such that  f(x0,y0,z0)f(x1,y1,z1).  Since D is connected, one can form the broken line P0Q1Q2QkP1 contained in D.  When one now goes along this broken line from P0 to P1, one mets the first corner where the value of f does not equal f(x0,y0,z0).  Thus D contains a line segmentMathworldPlanetmath, the end pointsPlanetmathPlanetmath of which give unequal values to f.  When necessary, we change the notations such that this line segment is P0P1.  Now, fx,fy,fz are continuousMathworldPlanetmathPlanetmath in D because they vanish.  The mean-value theorem for several variables guarantees an interior point(a,b,c)  of the segment such that


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