# vanishing of gradient in domain

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

 $\nabla{f}\;\equiv\;\vec{0}\quad\mbox{in}\;D,$

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

 $\displaystyle f_{x}^{\prime}(x,\,y,\,z)\;=\;f_{y}^{\prime}(x,\,y,\,z)\;=\;f_{z% }^{\prime}(x,\,y,\,z)\;=\;0\quad\mbox{for all}\;\;(x,\,y,\,z)\in D.$ (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  $f(x_{0},\,y_{0},\,z_{0})\neq f(x_{1},\,y_{1},\,z_{1})$.  Since $D$ is connected, one can form the broken line $P_{0}Q_{1}Q_{2}\ldots Q_{k}P_{1}$ 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

 $0\;\neq\;f(x_{1},\,y_{1},\,z_{1})-f(x_{0},\,y_{0},\,z_{0})\;=\;f_{x}^{\prime}(% a,\,b,\,c)(x_{1}\!-\!x_{0})+f_{y}^{\prime}(a,\,b,\,c)(y_{1}\!-\!y_{0})+f_{z}^{% \prime}(a,\,b,\,c)(z_{1}\!-\!z_{0}).$

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 VanishingOfGradientInDomain 2013-03-22 19:11:58 2013-03-22 19:11:58 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 26B12 partial derivatives vanish FundamentalTheoremOfIntegralCalculus ExtremumPointsOfFunctionOfSeveralVariables