Let \(W\subset\mathbb R^n\) be open and connected, and let \(f:W\to\mathbb R\) be a function all of whose partial derivatives vanish. Prove that \(f\) is constant.
Solution
1. For every \(p\in W\), there is an open box \(B\) containing \(p\)such that\[ B\subset W. \]
Since \(W\) is open, there is \(r>0\) such that the Euclidean ball \(B(p,r)\) is contained in \(W\). A sufficiently small open box centered at \(p\) is contained in this ball.
2. The function \(f\) is constant on every box \(B\) as in step
1.
Let \[ x=(x_1,\ldots,x_n), \qquad y=(y_1,\ldots,y_n) \] be points of \(B\). Define \[ z^{(j)} = (y_1,\ldots,y_j,x_{j+1},\ldots,x_n), \qquad 0\le j\le n, \] so that \(z^{(0)}=x\) and \(z^{(n)}=y\).
Because \(B\) is a box, the line segment from \(z^{(j-1)}\) to \(z^{(j)}\) lies in \(B\) and varies only in the \(j\)th coordinate. Restricting \(f\) to that segment gives a one-variable differentiable function whose derivative is \[ \frac{\partial f}{\partial x_j}=0. \] The mean value theorem therefore gives \[ f(z^{(j)})=f(z^{(j-1)}). \] Applying this successively for \(j=1,\ldots,n\) yields \[ f(y)=f(x). \] Thus \(f\) is constant on \(B\).
3. The function \(f\) is locally constant on \(W\).
For each \(p\in W\), step 1 supplies a neighborhood \(B\) of \(p\), and step 2 shows that \(f\) is constant on that neighborhood.
4. Fix \(p_0\in W\) and set\[ U = \{p\in W:f(p)=f(p_0)\}. \]Then both \(U\) and \(W\setminus U\) are open in \(W\).
If \(p\in U\), local constancy from step 3 gives a neighborhood of \(p\) on which every point has value \(f(p)=f(p_0)\); hence that neighborhood lies in \(U\). Thus \(U\) is open.
If \(p\in W\setminus U\), local constancy gives a neighborhood of \(p\) on which every point has value \(f(p)\neq f(p_0)\); hence that neighborhood lies in \(W\setminus U\). Thus the complement is also open.
5. One has\[ U=W. \]
The set \(U\) is nonempty because \(p_0\in U\). By step 4, \(U\) is both open and closed in the connected space \(W\). Therefore \(U=W\).
6. Consequently,\[ \boxed{f\text{ is constant on }W}. \]
By step 5, every \(p\in W\) satisfies \[ f(p)=f(p_0). \]
7. Q.E.D.
Step 6 is the required conclusion.