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If $f$ is a real function defined in a convex open set $E \subset R^{n}$, such that $\left(D_{1} f\right)(\mathbf{x})=0$ for every $\mathbf{x} \in E$, prove that $f(\mathbf{x})$ depends only on $x_{2}, \ldots, x_{n}$.
Show that the convexity of $E$ can be replaced by a weaker condition, but that some condition is required. For example, if $n=2$ and $E$ is shaped like a horseshoe, the statement may be false.
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We need to show that $f\left(x_{1}^{0}, x_{2}, \ldots, x_{n}\right)=f\left(x_{1}^{1}, x_{2}, \ldots, x_{n}\right)$ whenever $\mathbf{x}^{0}=\left(x_{1}^{0}, x_{2}, \ldots, x_{n}\right)$ and $\mathbf{x}^{1}=\left(x_{1}^{1}, x_{2}, \ldots, x_{n}\right)$ both belong to $E$. Since $E$ is convex, the line segment joining $\mathrm{x}^{0}$ and $\mathrm{x}^{1}$ is contained in $E$. The meanvalue theorem applies on this line segment, showing that $f\left(\mathbf{x}^{0}\right)-f\left(\mathbf{x}^{1}\right)=\left(x_{1}^{0}-\right.$ $\left.x_{1}^{1}\right)\left(D_{1} f\right)(\mathbf{x})$ for some point $\mathbf{x}$ on this interval. Hence the result now follows from the hypothesis.

Note that convexity is needed only on each line segment through $E$ parallel to the $x_{1}$-axis. Thus if the intersection of $E$ with each line parallel to the $x_{1}$-axis is an interval and $\left(D_{1} f\right)(\mathbf{x})=0$ for all $\mathbf{x} \in E$, then $f$ is independent of $x_{1}$.
If we define $f(x, y)$ on all of $R^{2}$ except the nonnegative portion of the $y$-axis by specifying
$f(x, y)= \begin{cases}0 & \text { if } y<0 \text { or } x<0, \\ y^{2} & \text { if } y \geq 0 \text { and } x>0,\end{cases}$
then $f(x, y)$ is continuously differentiable on its domain, $\left(D_{1} f\right)(x, y)=0$ everywhere on that domain, yet $f(-1,1)=0 \neq 1=f(1,1)$, so that $f$ is not independent of $x$.
by Diamond (66,887 points)

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