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Let $f:\mathbf{R}^k\to \mathbf{R}$ be a function such that all partial derivatives exists on all of $\mathbf{R}^k$.

If $D_i f(\vec{x})=0$ for all $\vec{x}\in\mathbf{R}^k$ and all $i$, show that there exists $c\in\mathbf{R}$ such that $f(\vec{x})=c$ for all $\vec{x}$.

Let $\vec{x}=(x_1,\ldots,x_k)\in\mathbf{R}^k$. Define for all $1\leqslant i\leqslant k$, $g_i:\mathbf{R}\to\mathbf{R}$ by $g_i(t)=f(x_1,\ldots,t,\ldots,x_k)$ with $t$ on the $i$-th place. Then $g'(x_i)=D_i f(\vec{x})=0$. Since $\vec{x}$ and thus $x_i$ is arbitrary, this implies that $g_i \equiv c_i\in\mathbf{R}$ according to some theorem in my textbook.

The thing is, how do I now use this information to conclude that $f$ itself is constant?

Community
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Dr. Heinz Doofenshmirtz
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3 Answers3

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If each $g_i$ is constant, then $f$ is constant. Indeed, take two points $(x_1,x_2,x_3),(y_1,y_2,y_3)\in\mathbf{R}^3$. Then\begin{align}f(x_1,x_2,x_3)&=f(y_1,x_2,x_3)\text{ because $g_1$ is constant}\\&=f(y_1,y_2,x_3)\text{ because $g_2$ is constant}\\&=f(y_1,y_2,y_3)\text{ because $g_3$ is constant.}\end{align}The same argument works for any number of variables.

Shuhul Mujoo
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José Carlos Santos
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Given $a\in{\mathbb R}^k$ consider the axis aligned box with opposite vertices $0$ and $a$. By assumption the function $f$ is constant along all edges of this box. It follows that $f(a)=f(0)$.

Your three lines after the highlighted text don't make sense to me.

Christian Blatter
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  • Nice geometric way to see it! +1 Although I disagree with your last comment: indeed, the restrictions $g_i$ of $f$ to the $i^{\textrm{th}}$ defined by OP being constant is exactly what you mean by "$f$ is constant along all edges of this box". – Prasun Biswas Mar 12 '18 at 18:46
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Note that

$$g'(x_i)=D_i f(\vec{x})=0\implies g_i \equiv c_i\in\mathbf{R} $$

is a consequence of MVT.

Applying the same to all the components it follows that $f$ itself is constant and by definition of function $f(\vec x)=c_i=c$.

user
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  • Thanks. I know that it is a consequence of the mean value theorem. But could you show a rigorous proof of $c_1=\cdots=c_k$? – Dr. Heinz Doofenshmirtz Mar 12 '18 at 15:01
  • Since f is constant with respect to each component we have that $c_i=c_j=c$ by definition of function. – user Mar 12 '18 at 15:05
  • Think to a simpler case $z=f(x,y)$. If $f(x,y_0)=c_1$ and $f(x_0,y)=c_2$ since $f(x_0,y_0)=c_1=c_2$ we have that $c_1=c_2=c$. – user Mar 12 '18 at 15:12
  • But each $g_i$ is defined by $\vec{x}$ right? So how do we know that $c_i$ is not dependent on $\vec{x}$? – Dr. Heinz Doofenshmirtz Mar 12 '18 at 15:29
  • @HeinzDoofenschmirtz The proof is in two step. Firstly by MVT we show that $f$ is constant in any restriction $g_i=c_i$ then, by definition of function $f=c_i=c$. The second part is trivial and maybe for that reason is not highlighted in the proof but it is a direct consequence of the defintion of function. See also the simple example given in the previous comment. – user Mar 12 '18 at 15:32