0
$\begingroup$

Let $\alpha \in\Bbb R$ and let $F:\Bbb R^k \to \Bbb R$ be positive homogenous function of degree $\alpha$ i.e $ \forall x \neq 0$ in $\Bbb R^k$ and for all $\lambda >0$ we have: $F(\lambda x)=\lambda^{\alpha}F(x)$ Assume that $F$ is continuously differentiable in $\Bbb R^{k}\setminus\{0\}$ and $F(x)>0$ $\forall x \in \Bbb R^{k}\setminus\{0\}\ $ Let $S=\{x| F(x)=1\}\ $

Now let $a \in S$ find the distance between the origin and the tangent space to the set $S$ at the point $a$.

What I did so far:

the tangent space to the set $S$ at the point $a$ is: $\{x\in\Bbb R^k|\mathbf \nabla F(a) \cdot \mathbf{(x-a)}=0\} $ so define $g(x) = \mathbf \nabla F(a) \cdot \mathbf{(x-a)}$ and $f(x)=|x|^2$ so I want to minimize $f$ with the constraint $g(x) =0$ using lagrange multipliers I have $ \nabla f(x) = \lambda \nabla g(x)$ now $\nabla g(x) = \nabla F(a)$ and $\nabla f(x) = 2(x_{1},x_{2},...,x_{k})$ so we have $2(x_{1},x_{2},...,x_{k})= \lambda \nabla F(a)\ $ we can use euler relation on $F$ and we have: $\ \mathbf{a} \cdot \mathbf{\nabla F(a)}=\mathbf{\alpha} \cdot \mathbf{F(a)}$. The point $a$ is in the set $S$ so we have $F(a)=1$. so euler relation become $\ \mathbf{a} \cdot \mathbf{\nabla F(a)}=\alpha$ puting all together we have: $$2(x_{1},x_{2},...,x_{k})= \lambda\alpha$$

I don't know how to continue from here. does my way untill now looks ok? if so can you give me a hint how I should continue?

$\endgroup$
2
  • $\begingroup$ Is there a reason we use calculus here instead of linear algebra? $\endgroup$
    – user469053
    Mar 23 at 19:23
  • $\begingroup$ I saw this question when I was studying multivariable calculus $\endgroup$ Mar 23 at 20:18

1 Answer 1

1
$\begingroup$

Let's assume $F(0)=0$. Let's also assume $\alpha\neq 0$, otherwise we can get arbitrarily close to the origin in $S$ by considering $F(ta)$ for arbitrarily small, positive $t$.

Fix $0\neq a\in S$. As you said, $$\underset{h\to 0}{\lim} \frac{F((1+h)a)-F(a)}{h}=a\cdot \nabla F(a).$$ But also, $$\underset{h\to 0}{\lim}\frac{F((1+h)a)-F(a)}{h}=F(a)\underset{h\to 0}{\lim} \frac{(1+h)^\alpha-1}{h}=\alpha F(a)=\alpha.$$ So $$a\cdot \nabla F(a)=\alpha\neq 0.$$ Let $N=\nabla F(a)/\|\nabla F(a)\|$, where $\|\nabla F(a)\|$ denotes the length of $\nabla F(a)$. Since $a\cdot \nabla F(a)\neq 0$, $\|\nabla F(a)\|\neq 0$, and we aren't dividing by zero.

Then $x\in S$ iff $x\cdot \nabla F(a)=\alpha$.

Any $x$ can be written as $$x=x_\parallel+x\perp,$$ where $x_\parallel=(x\cdot N)N$ and $x_\perp =x-x_\parallel$. Then $x_\parallel\perp x_\perp$, so $$\|x\|^2=\|x_\parallel\|^2+\|x_\perp\|^2.$$ As noted above, $x\in S$ iff $x\cdot \nabla F(a)=\alpha$ iff $x\cdot N=\alpha/\|\nabla F(a)\|$ iff $$x_\parallel=\frac{\alpha}{\|\nabla F(a)\|}N.$$ Minimizing $$\|x\|^2=\|x_\parallel\|^2+\|x_\perp\|^2$$ as $x_\perp$ ranges over $\{y:y\cdot N=0\}$ yields the choice $x_\perp=0$, and the minimizer is $$x=x_\parallel = \frac{\alpha}{\|\nabla F(a)\|} = \frac{\alpha}{\|\nabla F(a)\|^2}\nabla F(a).$$

Another way to think about this: $\nabla F(a)$ is a normal to the tangent space. Since $a$ is in the tangent space, the plane is $x\cdot \nabla F(a)=a\cdot \nabla F(a)=\alpha$. The vector in this plane closest to the origin is obtained by "traveling along the normal" $\nabla F(a)$ until we hit the plane. I put "traveling along the normal" because if $\alpha$ is negative, we will actually want to move in the opposite direction of $\nabla F(a)$ to hit the plane. In any case, if a plane has normal vector $F\neq 0$ and is defined by $x\cdot F=\alpha$, the minimum distance vector to the origin has the form $\lambda F$ for some $\lambda$ (because that's traveling along the normal) and must satisfy $$\lambda F\cdot F=\alpha,$$ or $$\lambda = \alpha/\|F\|^2.$$

$\endgroup$
2
  • $\begingroup$ You wrote: Any $x$ can be written as $$x=x_\parallel+x\perp,$$ where $x_\parallel=(x\cdot N)N$ and $x_\perp =x-x_\parallel$. I remember something like that when I was studying linear algebra,but that was a long time ago. Could you please direct me to notes or books that I could use to expand my knowledge on the subject? thank you $\endgroup$ Mar 24 at 11:45
  • 1
    $\begingroup$ For this situation, we don't need to know more about $x_\parallel$ and $x_\perp$, because we give their definitions in the comment. But in general, this is just the orthogonal projection onto the subspace spanned by $N$. But any search for "orthogonal projection" should give a good background about existence, definitions, and properties. Here is one such intro. textbooks.math.gatech.edu/ila/projections.html $\endgroup$
    – user469053
    Mar 25 at 9:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .