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Hi everyone this is a past exam question that I am studying as I go through my class that I am having trouble with, the full question is this:

Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be a differentiable function and suppose that

$$ \nabla f(\textbf{x}) = g(\textbf{x})\textbf{x} $$ for all $\textbf{x} \in \mathbb{R}^3$ where $g: \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function. Show that $f$ is constant on each sphere in $\mathbb{R}^3$ centered at $(0,0,0)$. i.e. show that if $\textbf{a}$, $\textbf{b}$ $\in \mathbb{R}^3$ and $\|\textbf{a}\| = \|\textbf{b}\|$ then $f(\textbf{a}) = f(\textbf{b})$.

I'm stuck on this question and not sure which direction to take. I'm really just throwing ideas around.

I do know that the gradient vector at $\textbf{a}$ is perpendicular to all the tangent lines to the level set $f^{-1}(c)$ at $\textbf{a}$.

Then letting $\textbf{x} = h(t) = (h_{1}(t), h_{2}(t), h_{3}(t))$ with $h(0)= \textbf{a}$.

I have $f(h(\textbf{0}))=c$. Then taking derivatives:

$$\frac{d}{dt}f(h(\textbf{0})) = \frac{d}{dt} c = 0$$

$$h'(0) \cdot \bigtriangledown f(h(\textbf{0})) = 0$$

Substituting in for $f(h(\textbf{0}))$ yields

$$h'(0) \cdot g(\textbf{a})\textbf{a} = 0$$

Now i'm really not sure where to go from here, how relate the magnitude of $\textbf{a}$ or $\textbf{b}$ to the question (do i need to parametrise?) I would really appreciate some guidance with this question.

Cortizol
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JackReacher
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  • As written, the claim is false. Let $g(x)$ be the function that returns the first coordinate of $x$. Then $g(x)x$ is differentiable and always radially directed, except on the $y$-$z$ plane where it is zero, so is not radially directed (it's still differentiable there). $f$ isn't constant on any sphere of positive radius. Are you sure there isn't an additional condition? – Eric Towers Mar 19 '14 at 10:16
  • @EricTowers - Sorry I'm not sure what you mean by let $g(x)$ be the function that returns the first coordinate of $x$. The question states the $g(x)$ is $f:R^3 \rightarrow R$? – JackReacher Mar 19 '14 at 10:32
  • The problem does not say this about $g$. The problem only requires that $g$ is a function with domain $\mathbb{R}^3$ and range $\mathbb{R}$. There are no further restrictions (stated) on $g$. The function $g((x_1,x_2,x_3)) \mapsto x_1$ certainly meets this prescription. – Eric Towers Mar 19 '14 at 10:34
  • But then what does $g(\textbf{x})\textbf{x}$ mean? $\textbf{x}$ is a vector. My understanding is that $g(\textbf{x})$ just returns a constant, multiplied to the vector $\textbf{x}$. I checked the question, and I haven't missed anything. – JackReacher Mar 19 '14 at 10:40
  • Yes. "$g(x)x$" is the vector in the direction of $x$ with magnitude $g(x)|x|$. There is nothing in the above requiring that $g(x)$ is constant for all $x$. – Eric Towers Mar 19 '14 at 10:41
  • I'm sorry, I appreciate your help, but I am still not understanding. The way I read the question is that the gradient vector at any point is a scalar multiple of the vector giving that point. Since $g(\textbf{x})$ is from $R^3 \rightarrow R$, so it just returns a number/constant right? What else could it return? What could the question be missing? – JackReacher Mar 19 '14 at 10:53
  • $g(x)$ returns a scalar. That scalar can be different for every $x$. In fact it would be just as good to write $\nabla f(x) = \langle\text{a randomly selected number for each }x\rangle x$ and then hope that this somehow maintains spherical symmetry. – Eric Towers Mar 19 '14 at 10:56
  • let us continue this discussion in chat – JackReacher Mar 19 '14 at 10:59
  • Can't. Heading to bed. Good luck. – Eric Towers Mar 19 '14 at 11:00
  • You wish to prove that $f$ is a radially symmetric function, namely a function that depends only on $r=|x|$: $f(x)=\tilde{f}(r)$. But for such a function (assuming differentiability) $\nabla f(x)=\tilde{f}'(r)r$ as you can easily check by the chain rule. So you have a necessary condition that looks rather different than your assumption on $g$. – Siminore Mar 19 '14 at 11:17
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    @EricTowers your proposed counterexample does not work. With this choice of $g$, $g(x)x$ is not the gradient of a function. – Steven Gubkin Mar 19 '14 at 11:19
  • @Siminore I understand your first two sentences, so then I would have $\tilde{f}'(r)r = g(\textbf{x})\textbf{x}$. So then what does this mean? Could you elaborate a little further for me? Thanks! – JackReacher Mar 19 '14 at 21:09

2 Answers2

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Pick a path $\gamma:[0,1]\to \mathbb{R}^3$ with $\gamma(0) = a$ and $\gamma(1) = b$, and with $|\gamma(t)| = r = |a|=|b|$ for all $t$, i.e. $\gamma$ lies entirely on the sphere of radius $r= |a|=|b|$ centered at the origin.

I want to show that $(f \circ \gamma): [0,1] \to \mathbb{R}$ is a constant function. One way to do this is to show that its derivative is always $0$.

$\begin{align*}(f \circ \gamma)'(t) &= \nabla f(\gamma(t)) \cdot \gamma'(t) \text{ by the chain rule} \\ &=g(\gamma(t))\gamma(t) \cdot \gamma'(t)\\ &=0 \end{align*}$

The last equality follows since $\gamma(t)$ is a radial vector and $\gamma'(t)$ is a tangent vector to the sphere. Another way to see this is to use the chain rule again:

Let $R(x,y,z) = x^2+y^2+z^2$. Then $R(\gamma(t)) = r$ for all $t \in [0,1]$. By the chain rule, $$ \nabla R(\gamma(t))\cdot \gamma'(t) = 0 $$

But by a computation, $\nabla R (x,y,z)= 2\begin{bmatrix}x\\y\\z\end{bmatrix}$. In other words $\nabla R (v) = 2v$ So we have

$$ 2\gamma(t)\cdot \gamma'(t)=0 $$

Thus $(f \circ \gamma)'(t) = 0$ for all $t \in [0,1]$, so $(f \circ \gamma)$ is constant. So $f(a) = f(b)$. Since $a$ and $b$ were arbitrary points on the sphere of radius $r$, we have that $f$ is constant on any sphere centered at the origin.

Steven Gubkin
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  • sorry did you mean that $\gamma (t)$ is perpendicular to the sphere? Also, (sorry this might be a dumb question, but why do you have that $\gamma$ has domain [0,1]? Many thanks for your help! – JackReacher Mar 19 '14 at 23:48
  • Paths usually have domain $[0,1]$, but this is convention: they have to start and end somewhere! $\gamma{t}$ is a point on the sphere, or a vector from the origin to the sphere depending on how you look at it. As a vector it is always perpendicular to tangent vectors to the sphere (a radial line to a sphere is perpendicular to the tangent plane at that point) – Steven Gubkin Mar 19 '14 at 23:58
  • Thanks, so just to clarify, in your answer, where you have '$\gamma '(t)$ is perpendicular to the sphere', you actually meant '$\gamma (t)$ is perpendicular to the sphere' Is that correct? Also, where you have $f:[0,1] \rightarrow R$ that is analogous to $f: R^3 \rightarrow R$? – JackReacher Mar 20 '14 at 01:44
  • Correct, sorry. – Steven Gubkin Mar 20 '14 at 10:45
  • I typed this answer as it took form in my mind, but I think it might be clearer if I rework it. Going to make a major edit. – Steven Gubkin Mar 20 '14 at 10:49
  • @mathstudent What do you not like about this answer exactly? – Steven Gubkin Mar 21 '14 at 12:59
  • I just wanted to see if there were additional ways to answer the question - just to expand my intuitive understanding. Your answer is very good though, thank you. – JackReacher Mar 22 '14 at 05:49
  • You said you wanted an answer with more rigor. I do not see what is not rigorous about my proof. The only thing (maybe) is that $\gamma(t)$ is perpendicular to $\gamma'(t)$. The proof of this is just another application of the chain rule. Would you like me to put that in? – Steven Gubkin Mar 22 '14 at 13:13
  • Yes that would be good. Thanks for all your help. – JackReacher Mar 24 '14 at 06:11
  • Thanks for your help Steven, I do have a question: While I see clearly that you have demonstrated that if two vectors are of the same length then, we do indeed get a constant function, but is this the same as showing $f(a)=f(b)$? Could it be that $f(a)$ and $f(b)$ both given constants, but different constants? – JackReacher Mar 26 '14 at 08:43
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Here is the idea:

If $|\mathbf{a}| = |\mathbf{b}|$, pick a path $\gamma$ lying entirely within the sphere of that common radius connecting $\mathbf{a}$ and $\mathbf{b}$

By the fundamental theorem of calculus,

$\begin{align*} f(\mathbf{b}) - f(\mathbf{a}) &= \int_\gamma \nabla f \cdot \vec{dr}\\ &=\int_\gamma g(x)\vec{x}\cdot \vec{dr} \end{align*}$

You need to conclude that this integral is $0$. Can you see why it is?

Steven Gubkin
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  • We actually haven't gone into the integration part of the course yet, so I'll have to look ahead a bit..is there a way to do it using differentiation? Thanks. – JackReacher Mar 19 '14 at 21:14