Show that there is no continuously differentiable function on closed disks

Question

In Cartesian coordinates, the closed unit disk is given by $$ \bar{D}=\left\{(x, y) \mid x^2+y^2 \leq 1\right\}, $$ and its boundary is given by $$ \partial D=\left\{(x, y) \mid x^2+y^2=1\right\}. $$ Prove that there is no continuously differentiable function $f: \bar{D} \rightarrow \mathbb{R}^2 $ satisfying: $$ f\left(\bar{D}\right) \subseteq \partial D \;\;\text{and}\;\; f(x,y)=f\left(x y, x^2-y^2\right). $$ Here, I have no idea why the question emphasize $f$ need to be differentiable. If someone can help me understand how to think about it would be great.

Answer 1

Let's note $g : (x,y) \mapsto (xy,x^2-y^2)$ ; we therefore have $f = f \circ g$. Four points here :

1. Where does this function $g$ come from ? We investigate and wonder if this is the right $g$ to consider.

2. There actually are some functions that satisfy your hypothesis : any function $f$ which is constant on $\overline{D}$ and whose (only) value is some point in $\partial D$ is a solution to your problem.

3. These are actually the only ones that satisfy the required property. I'll give a proof without using diffentiabilty.

4. How to use differential here : some ideas but nothing conclusive.

Let's go !

1. Here, one can see that every point $(x,y) \in \overline{D}$ can be written as $(x,y) = (r\cos(\theta),r\sin(\theta))$ with $r \in [0,1]$. There, we get $$g(x,y) = (xy, x^2-y^2) = (r^2 \cos(\theta)\sin(\theta), r^2(\cos^2(\theta)-\sin^2(\theta)) = r^2 (\frac{1}{2}\sin(2\theta),\cos(2\theta)).$$

This makes me wonder if the more interesting property to look at may rather be $$f(x,y) = f(2xy,x^2-y^2)$$ so that we get $$f(r\cos(\theta),r\sin(\theta)) = f(r^2\sin(2\theta),r^2\cos(2\theta))$$ which would greatly ease our study.

2. This one is self-explanatory.

3. Let $f$ be a function satisfying our assumptions. We will show that $f$ is constant on $D$. As $f$ is continuous, it will be enough to say that $f$ is constant on $\overline{D}$.

Let $p =(r\cos(\theta),r\sin(\theta)) \in D$ : we have $ r \in ]0,1[$. We've seen that $g(p) = r^2 (\frac{1}{2}\sin(2\theta),\cos(2\theta))$, therefore $$\|g(p)\| \leq r^2 = \|p\|^2.$$ Let's consider the sequence $(p_n)_{n \in \mathbb{N}}$ of iterates of $p$ by $g$, i.e. the sequence defined by $p_0=p$ and $p_{n+1}=g(p_n)$ for all $n \in \mathbb{N}$. This is relevant because, as $f \circ g = f$, we get $f(p_{n+1})=f_(p_n)$ for all $n$, and therefore $f(p_n)=f(p)$ for all $n$.

With what we saw on $g$, we have $\| p_{n+1} \| \leq \|p_n\|^2$ and by induction we obtain $\|p_n\| \leq \|p\|^{2^n}$. As $\| p \|<1$, this gives $\|p_n\| \xrightarrow[n \to +\infty]{} 0$, and therefore $p_n \xrightarrow[n \to +\infty]{} (0,0)$. As $f$ is continuous, we get $$f(p_n) \xrightarrow[n \to +\infty]{} (0,0)$$ hence $$f(p)= f((0,0)).$$ This is true for all $p \in D$, therefore $f$ is constant on $D$.

4. The hypothesis "$f$ differentiable" invites us to differentiate the relation $f =f \circ g$.

$$\forall p \in \overline{D}, \forall u \in \mathbb{R}^2,\quad df_{p}[u] = df_{g(p)}[dg_p[u]].$$

One can compute $dg_p$ for $p=(x,y)$ and see that the matrix of this differential is $\begin{pmatrix}y& x \\ 2x&-2y\end{pmatrix}$. In particular, as $\begin{pmatrix}y& x \\ 2x&-2y\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix} =\begin{pmatrix}2xy \\ 2x^2-2y^2\end{pmatrix}$, we have $dg_p[p]= g(p)$ and therefore : $$df_p(p)=2.df_{g(p)}[g(p)].$$ But the only way I see to use that is to do a similar reasoning as in point 2... Once again, this would be simpler if we had the property highlighted in point 1.

$\rhd$ As $f(\overline(D))\subset \partial D$, we know that the norm of of $f$ is constant equal to 1, i.e. $p \mapsto \langle f(p) ,f(p) \rangle $ is constant equal to 1. Therefore, the differential of this function is zero, i.e. : $$\forall p \in \overline{D}, \forall u \in \mathbb{R}^2, \quad 2 \langle f(p) ,df_p[u] \rangle =0, $$ thus $df_p[u] $ is orthogonal to $f(p)$. One may try to combine this with the previous computation to find something relevant : for now, I didn't find a meaningful equality.

Answer 2

The exercise has been remembered with a few mistakes. The formula given for $f$ should describe $f\vert_{\partial D}$, and it should be corrected to $f(x,y)=(2xy, x^2- y^2)$ One should prove that there is no continuous $f:\overline{D}\to \partial D$ with this boundary condition. A forteriori there is no continuously differentiable map with this property either. Depending on the context where this exercise occurs the stronger condition on f allows methods of proof that would not directly work for the continuous case. Saying more would spoil the fun, I'm afraid. (If you know what the winding number is, that would be one way to go.)