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# vanishing of gradient in domain

Theorem. If the function $f$ is defined in a domain $D$ of $\mathbb{R}^{n}$ and all the partial derivatives of a $f$ vanish identically in $D$, i.e.

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

*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.

## Mathematics Subject Classification

26B12*no label found*

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