1

Suppose $r>0$, $a\in\mathbb{R}^n$, and $f\colon B_r(a)\to\mathbb{R}^m$. If all first order partial derivatives of $f$ exist on $B_r (a)$, and $f_{x_j}(x) = 0$ for all $x\in B_r (a)$ and all $j=1,2,\ldots ,n$. Prove $f$ has only one value on $B_r(a)$.

I guess I don't quite understand the problem so I can't solve it. If there is a ball of radius $r>0$ centered at $a\in\mathbb{R}^n$, its not clear to me how $f$ can only have one value on $B_r(a)$. Also, what does every partial derivative being equal to $0$ for all $x\in B_r(a)$ tell me?

Can anyone give me some hints and/or solve this problem?

Git Gud
  • 31,356
durantula35
  • 11
  • Consider the bidimensional version of the result. You know that $\frac{\partial f}{\partial x_1}$ is always $0$, integrating with respect to the first coordinate tells you that $f(x_1,x_2)=\varphi(x_2)$ for a certain differentiable one-dimensional function $\varphi$ and $(x_1, x_2)$ in the disc. Now differentiate the equality with respect to the second parameter to conclude. – Git Gud Feb 16 '15 at 22:54
  • Can you explain how the integral with respect to x1 of ∂f/∂x1 (which is equal to zero) gives a function of of only x2? Shouldn't the integral be 0, plus a some constant? – durantula35 Feb 18 '15 at 01:20
  • In analyst's slang, it is a constant with respect to the first variable, but not necessarily to the second variable. When you differentiate with respect to a variable, all others act like constants, the same principle applies for integration. – Git Gud Feb 18 '15 at 11:07
  • See also this. – Git Gud Feb 18 '15 at 11:14

0 Answers0