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Let $u:[0, 1]\to\mathbb R$ be of class $C^1$ over $[0, T]$. As an exercise I have to determine the boundary of the set

$$ A=\{u : |u^{\prime}|\le 2 \text{ and } u(x)\neq k\pi,\forall k\in\mathbb{Z}, \forall x\in [0, 1] \}.$$

Here is considered on $C^1([0, 1])$ the topology coming from $\| u\|_\infty +\| u'\|_\infty$.

I am in trouble in determining the boundary of $A$. If $k=0$, I note that $A$ reduce to

$$ A=\{u : |u^{\prime}|\le 2 \text{ and } u(x)\neq 0, \ \forall x\in [0, 1] \}$$ and its boundary should be

$$ \partial A=\{u : |u^{\prime}|\le 2 \text{ and } u(x)=0, \ \forall x\in [0, 1] \}.$$

I don't know how to proceed in the case $k\neq 0$. Could someone please help?

Thank you in advance.

Arctic Char
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  • In the case you mention ($k=0$), it seems to me that functions $u: |u^{\prime}|\le 2 \text{ and } \exists x \in [0,1], u(x) = 0$ belong to $\partial A$, too. – Jean-Armand Moroni Apr 08 '23 at 21:46
  • Do you mean that the norm is $|u|\infty$ or $|u|\infty+|u’|_\infty$? In the first case, you are going to destroy the condition $|u’|\le2$. – Gio67 Apr 10 '23 at 09:27
  • @Gio67, the norm is $|u|\infty +|u'|\infty$. –  Apr 10 '23 at 12:02

1 Answers1

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The way you interpret $k$ is a bit off: the condition is that

$$ u(x) \neq k\pi$$

for all $k\in \mathbb Z$ and $x\in [0,1]$. So you cannot split into two cases ($k=0$ and $k\neq 0$).

If $u\in A$, by the intermediate value theorem,

$$ u \in A_k = \{ u\in A : k\pi < u(x) < (k+1) \pi, \ \ \forall x\in [0,1]\}$$

for some $k\in \mathbb Z$, and $A$ is the disjoint union of $A_k$ for $k\in \mathbb Z$.

It is easy to show that the closure of $A_k$ is

$$\overline {A_k} = \{ u\in C^1[0,1] : |u'|\le 2, \ \ k\pi \le u(x) \le (k+1) \pi\}$$

By definition of $\partial A$, we need to find the interior $A^\circ $ of $A$. We claim that

$$\tag{1} A^\circ = \{ u\in A : |u'(x)| <2, \ \ u(x) \neq k\pi,\ \ \forall k\in \mathbb Z, x\in [0,1]\}.$$

The inclusion $\supset$ is clear. On the other hand, if $u\in A$ and is not in the right hand side of $(1)$. Then there is $x_0\in [0,1]$ so that $u'(x_0) = \pm 2$. Then for all $\epsilon >0$,

$$ u_\epsilon (x) = u(x) \pm \epsilon x$$

satisfies $u'_\epsilon (x_0) = \pm (2+\epsilon)$ Hence $u_\epsilon \notin A$. But clearly $\| u_\epsilon - u\| \le 2\epsilon$. Hence $u$ is not in $A^\circ$.

Finally, by definition of $\partial A = \overline A \setminus A^\circ$,

\begin{split} \partial A &= \{ u : |u'|\le 2,\ k\pi \le u(x) \le (k+1)\pi \ \ \forall x\in [0,1], \text{ for some } k\in \mathbb Z \\ &\ \ \ \ \ \ \ \ \text{and } \exists x_0\in[0,1] \text{ such that either } \\ &\ \ \ \ \ \ \ \ u'(x_0) = \pm 2 \text{ or } u(x_0) = k\pi \text{ or } u(x_0) = (k+1)\pi\} \end{split}

The description is a bit complicate, but for example, $\partial A$ contains infinitely many constant functions $u(x) = k\pi$ (for all $k\in \mathbb Z$), and it also contains (e.g.) $u(x) = x^2$ and $u(x) = 2x$.

Arctic Char
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    $u(x_0)=2$ should be $u’$. Also, you should still keep $|u’|\le 2$ in $\partial A$ – Gio67 Apr 10 '23 at 13:27
  • Thanks, it's fixed. @Gio67 – Arctic Char Apr 10 '23 at 13:29
  • @ArcticChar thank you for the answer. I don't understand where does the condition "$\exists x_0\in [0, 1]$ such that..." come from in the definition of $\partial A$. Could you please specify it? thank you. –  Apr 10 '23 at 20:12
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    Elements in $\overline A_k$ are of the form $|u'|\le 2$ and $k \pi \le u\le (k+1)\pi$, and the interior $A_k^\circ$ contains $u$ of the form $|u'(x)|<2$ and $k\pi < u(x) < (k+1) \pi$ for all $x$. Thus the difference $\overline A_k \setminus A^\circ_k$ contains $u$ of the form $|u'(x)|=\pm 2$ or $k\pi = u(x)$ or $u(x) = (k+1) \pi$ for some $x$. I am more or less just negating the statement "$|u'(x)|<2$ and $k\pi < u(x) < (k+1) \pi$ for all $x$", @C.Bishop – Arctic Char Apr 11 '23 at 05:57