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.