vanishing of gradient in domain
then the function has a constant value in the whole domain.
Proof. For the sake of simpler notations, think that ; thus we have
Make the antithesis that there are the points and of such that . Since is connected, one can form the broken line contained in . When one now goes along this broken line from to , one mets the first corner where the value of does not equal . Thus contains a line segment, the end points of which give unequal values to . When necessary, we change the notations such that this line segment is . Now, are continuous in because they vanish. The mean-value theorem for several variables guarantees an interior point 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|
|Date of creation||2013-03-22 19:11:58|
|Last modified on||2013-03-22 19:11:58|
|Last modified by||pahio (2872)|
|Synonym||partial derivatives vanish|