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If $f$ is a real function defined in a convex open set $E\subset \mathbb{R}^n$, such that $(D_1f)(\mathbf{x})=0$ for every $\mathbf{x}\in E$, prove that $f(\mathbf{x})$ depends only on $x_2, \dots, x_n$.

Proof: Let $\overline{a}\in E$ and $\hat{a}\in E$ two points which differs only by first coordinate, i.e. $\overline{a}=(a_1,a_2,\dots, a_n)$ and $\hat{a}=(b_1,a_2,\dots, a_n)$ and WLOG we suppose that $a_1>b_1$. We have to prove that $f(\overline{a})=f(\hat{a})$.

Let $f(t,x_2,\dots,x_n)=g(t)$ then $$f(\overline{a})-f(\hat{a})=f(a_1,a_2,\dots, a_n)-f(b_1,a_2,\dots, a_n)=g(a_1)-g(b_1)=$$ Here we can use Mean value theorem and we get: $$=g'(\theta)(a_1-b_1)$$ where $\theta\in (a_1,b_1)$. But $(\theta, a_2,\dots,a_n)\in E$ since $E$ is convex. Then $g'(\theta)=(D_1f)(\theta)=0$ $\Rightarrow$$g(a_1)-g(b_1)=0$. Thus, $f(\overline{a})=f(\hat{a})$.

I am sorry if this topic is repeated but I solved this problem after many attempts.

Is this proof correct?

RFZ
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    This proof looks fine. You should probably mention why $g(t)$ satisfies the conditions of the Mean value theorem. – Oliver Jones Feb 12 '16 at 23:17
  • @OliverJones, The Mean value theorem is applicable since $g(t)$ is differentiable at $[b_1,a_1]$ because for any $u\in [b_1,a_1]$ we get $(u,a_2,\dots,a_n)\in E$ and $(D_1f)(u, a_2, \dots, a_n)=0.$ Am I right? – RFZ Feb 14 '16 at 20:11
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    You need $g$ to be differentiable on your interval $(b_1, a_1)$ but this assured by the existence of $D_1f$ on $E$. You don't need $(D_1f)(u,a_2,\cdots ,a_n)=0$. – Oliver Jones Feb 14 '16 at 23:56
  • @OliverJones, Yes you are right! $g$ must be differentiable on interval $(b_1,a_1)$ and I'm not need at $(D_1f)(u,a_2, \dots, a_n)=0$ ( I just wrote it). Thank you very much for your remark! – RFZ Feb 15 '16 at 05:33

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