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Problem saying that :

$f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is differentiable. Assume that there is a differentiable function $g:\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $\nabla f(x)=g(x)x$ . Show that $f$ is constant on $S=\{x\in\mathbb{R}^{n}:||x||=r\}$ where $r$ is positive constant.

For $x=(x_1,\dots ,x_n)$ and $\nabla f=(\frac{\partial f}{\partial x_{1}},\dots,\frac{\partial f}{\partial x_{n}})$, problem says $\frac{\partial f}{\partial x_{i}}=g(x)x_{i}$. It seems to me that to solve this problem, knowing the relation between norm of gradient and its value is crucial. How can I do?

Notification : This question is edited since it's about same problem and the former one is about just notation.

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    $\begingroup$ g(x) is a scalar. $\endgroup$
    – Samuel
    Commented Mar 25, 2016 at 12:25
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    $\begingroup$ For every $x$, $g(x)$ is a scalar. You can rightfully multiply scalars by vectors, in this case it's the scalar $g(x)$ times the vector $x$. $\endgroup$
    – Git Gud
    Commented Mar 25, 2016 at 12:26
  • $\begingroup$ I see. $g$ is a real-valued.... How could I delete this question or replace it about the way to solve that problem? $\endgroup$
    – Darae-Uri
    Commented Mar 25, 2016 at 12:28
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    $\begingroup$ Hint for your modified question: First show $\nabla \|x\| = \dfrac{x}{\|x\|}$, then use the chain rule with the ansatz $f(x) = \phi(\|x\|)$ for some real-valued function $\phi$ of one variable. $\endgroup$ Commented Mar 25, 2016 at 12:36
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    $\begingroup$ To add some intuition to what @AndrewD.Hwang said: The assumption says that $\nabla f$ points in the radial direction, which means that $f$ will not change when you move along a sphere centered at the origin. So $f$ should be a function of $\|x\|$ alone. $\endgroup$ Commented Mar 25, 2016 at 12:43

1 Answer 1

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Note that $f$ is constant on a sphere $S$ of radius $r$ iff for any curve $c :(-\epsilon,\epsilon)\rightarrow S$, $f\circ c(t)=C$ for all $t$ iff $ \frac{d}{dt} (f\circ c)(t)=0$

If $c$ is a curve on the sphere so $|c(t)|=r$ Then $$ c'(t)\cdot c(t)=0 $$

Hence \begin{align} \frac{d}{dt} f\circ c(t) &=\nabla f\cdot c' \\ &= g(c(t))c(t)\cdot c'(t) =0 \end{align}

That is $f$ is constant on the curve.

(And note that gradient of $f$ is radial direction That is hypersurface, which is level surface of $f$, has unit normal of radial direction. Such kind of hypersurface is a sphere.)

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  • $\begingroup$ How can r be parametrized by a single variable t? $\endgroup$
    – Mathcho
    Commented Mar 25, 2016 at 14:37
  • $\begingroup$ I add more explanation $\endgroup$
    – HK Lee
    Commented Mar 25, 2016 at 14:42
  • $\begingroup$ How do I show the equivalence between f is constant on S and f is constant on such curves? $\endgroup$
    – Darae-Uri
    Commented Mar 27, 2016 at 5:50
  • $\begingroup$ Please read again. "for any curve $c$ ..." $\endgroup$
    – HK Lee
    Commented Mar 27, 2016 at 5:51

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