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Let $f:\mathbb{R}^n\to\mathbb{R}^n$ given by $f(x)=|x|^2x$. (a) Show that $f\in\mathcal{C}^{\infty}$ and carries the unity open ball to itself in an one-to-one fashion. (b) Calculate the Jacobian matrix of $f'(x)$. (c) Show that $f$ carries the ball centered at $e_1=(0,\dots,0)$ to an open set. (d) Show that the inverse of $f$ is not differentiable at $0 = (0,\dots, 0)$.

I'm starting by calculating the partial derivates to calculate the Jacobian and to show continually differentiability. Let $x = (a_1,\dots,a_n)$. Then (skiping lots of tedious calculations)

$$x+te_i=(a_1,\dots,a_i+t,\dots,a_n),$$ $$f(x+te_i) = (a_1^2+\dots+(a_i+t)^2+\dots+a_n^2)(a_1,\dots,a_i+t,\dots,a_n)$$ $$f(x+te_i)-f(x) = (a_1((a_i+t)^2-a_i^2),\dots,ta_i(2a_i+t)+t(a_1+\dots+(a_i+t)^2+\cdots+a_n),\dots,a_n((a_i+t)^2-a_i^2)$$ $$\frac{f(x+te_i)-f(x)}{t} = (a_1(2a_i+t),\dots,a_i(2a_i+t)+(a_1+\dots+(a_i+t)^2+\cdots+a_n),\dots,a_n(2a_i+t))\to (2a_ia_1,\dots,a_1^2+\cdots+3a_i^2+\cdots+a_n^2,\dots2a_ia_n).$$

Is this correct?

If so, we know that the partials of $f$ exist and are continuous, so $f\in\mathcal{C}^1$. But I don't know exatly how to compute the Jacobian of $f'$. And assuming $f\in\mathcal{C}^n$, I don't know how to prove that it is $\mathcal{C}^{n+1}$. Any help with these or any of the other items?

Ain't No O
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    Proving $f$ is $C^{\infty}$ here is actually very simple because if you write $f(x) = (f_1(x), \dots, f_n(x))$, then it is easy to see that each component function $f_i(x)$ is a polynomial function in the variables $x_1, \dots, x_n$ (of order $3$). So, all the partial derivatives, to all orders exist and are continuous (in fact, all the 4th order partial derivatives are identically $0$). – peek-a-boo Jun 26 '20 at 04:31
  • @peek-a-boo Oh yes, I think I made it too complicated but now I got it. Could you please take a look at the other parts? For the Jacobian of $f'$, because it is a function $f':\mathbb{R}^n\to\mathcal{L}(\mathbb{R},\mathbb{R})$, how will I compute it? – Ain't No O Jun 26 '20 at 05:14
  • The derivative of $f'$ is $f''$, which can be thought of as a map $\Bbb{R}^n \to \mathcal{L}^2(\Bbb{R}^n,\Bbb{R}^n)$ (the space of bilinear maps $\Bbb{R}^n \times \Bbb{R}^n \to \Bbb{R}^n$). So as you can see, it gets slightly messy. For a start, I'd suggest you take a look at this answer of mine to see how to think about higher derivatives. – peek-a-boo Jun 26 '20 at 05:33

1 Answers1

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You can just use the definitions (details are left to you).

For all $x,h\in\mathbb{R}^n$, we have $f(x+h)=\vert x+h\vert^2(x+h)=(\vert x\vert ^2 +2(x\cdot h)+\vert h\vert^2)(x+h)=f(x)+(\vert x\vert^2 h+2(x\cdot h)x)+o(\vert h\vert ).$ Thus $f'(x)(h)=\vert x\vert^2 h+2(x\cdot h)x$.

Now for all $h,k$, $f'(x+k)(h)=(\vert x\vert^2+2 x\cdot k +\vert k\vert^2) h+2(x\cdot h+k\cdot h))(x+k)=f'(x)(h)+2(x\cdot k)h+ 2(x.h) k+ 2 (k\cdot h) x+\mathrm{negligible \ terms}$.

Hence $f''(x)(h,k)=2(x\cdot k)h+ 2(x.h) k+ 2 (k\cdot h) x$. If $(h,k)=(e_i,e_j)$, we then get $\dfrac{\partial^2f}{\partial x_i\partial x_j}(x)=2x_j e_i+2x_i e_j+\delta_{ij} x.$

Note that $f''(x)(h,k)$ is linear in $x$, so it is differentiable and $f'''(x)(h,k,\ell)=f''(\ell)(h,k)=2(\ell\cdot k)h+ 2(\ell.h) k+ 2 (k\cdot h) \ell$.

Now easily $f^{(4)}(x)=0$.

GreginGre
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