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Let $\mathcal{C}^1([a,b],\mathbb{R})$ be the vectorial space of functions of class $C^1$ (i.e. functions $f:[a,b]\to\mathbb{R}$ such that $f'$ is continuous) with a norm given by

$$||f||_1=\sup_{x\in[a,b]}\left(|f(x)|+|f'(x)|\right).$$

Let $H\subset\mathcal{C}^1([a,b],\mathbb{R})$ the subset of injective functions prove that if $h\in H$ and there exists $x\in[a,b]$ such that $h'(x)=0$ $h\notin \operatorname{int}(H)$. In particular $H$ is not open.

Attempt:

If $h\in H$ then is a continuous injection therefore $h$ is a homeomorphism over some close interval. Given $\epsilon>0$ exist $f\in B(h,\epsilon)$ such that $f'=0$.

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2 Answers

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Suppose $h \in H$ such that $\exists x_0$ with $h'(x_0) = 0$. Since $h$ is injective, it must be monotonic. Hence $h'$ must either be $\geq 0$, or $\leq 0$. I will assume that it is decreasing, the other case is handled similarly. Note that since $h$ is decreasing, we have $h(a) > h(b)$.

Let $\epsilon >0$, and define $h_{\epsilon}$ as $h_{\epsilon}(x) = h(x) + \epsilon (x-x_0)$. Clearly $h_{\epsilon}'(x) = h'(x) + \epsilon$, and we have $||h-h_{\epsilon}||_1 \leq \epsilon \max(|a-x_0|,|b-x_0|)+\epsilon$, so $h_{\epsilon}$ can be chosen arbitrarily close to $h$.

Suppose $x_0 \in (a,b)$. Then we have $h(a) > h(x_0) > h(b)$. Now choose $\epsilon$ small enough so that $h_{\epsilon}(a) > h_{\epsilon}(x_0) > h_{\epsilon}(b)$. Since $h_{\epsilon}'(x_0) = \epsilon >0$, we can conclude that there must be a point $x_1 \in (a,x_0)$ such that $h_{\epsilon}(x_1) = h_{\epsilon}(x_0)$, which means that $h_{\epsilon} \notin H$. Hence $h\notin H^{\circ}$.

If $x_0 = a$, choose $\epsilon$ small enough so that $h_{\epsilon}(a) = h_{\epsilon}(x_0) > h_{\epsilon}(b)$. By similar reasoning, we see that there must be a point $x_1 \in (x_0,b)$ such that $h_{\epsilon}(x_1) = h_{\epsilon}(x_0)$, with the same consequences.

The case $x_0=b$ is handled similarly.

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Nice construction! Thanks!! –  Gastón Burrull May 26 '12 at 21:14
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Suppose that given $h\in H$, there exists some $x\in [a,b]$ such that $h'(x)=0$.

Because $h$ is injective, we cannot have $h'\equiv 0$ on any open interval, so there is some open interval $(c,d)\subset [a,b]$ containing $x$ such that $|h'(t)|>0$ for all $t\in(c,d)$, $t\neq x$. Because $h'$ is continuous, $h'(t)$ has the same sign for all $t\in(c,x)$.

Because $h'$ is continuous and $h'(x)=0$, for any $\epsilon>0$ we can choose our interval $(c,d)$ containing $x$ to also have the property that $|h'(t)|<\frac{\epsilon}{4}$ for all $t\in (c,d)$.

For any $r,s$ with $c<r<x<s<d$, we can create a non-zero $C^\infty$ function $w:[a,b]\to\mathbb{R}$ supported on $[r,s]\subset (c,d)$ such that

  • $\|w\|<\epsilon$
  • $|w'(z)|>\frac{\epsilon}{2}$ for some $z\in (r,x)$
  • $w'$ has the opposite sign of $h'$ on $(r,x)$

Thus $$\|(h+w)-h\|=\|w\|<\epsilon,$$ and $h+w\in C^1([a,b],\mathbb{R})$, but the sign of $(h+w)'$ changes between $r$, where $$(h+w)'(r)=h'(r)+w'(r)=h'(r)+0=h'(r)$$ has the same sign as $h'$, and $z$, where $$(h+w)'(z)=h'(z)+w'(z)$$ has the opposite sign of $h'$ because $w'$ has the opposite sign of $h'$ on $(r,x)$ and $|w'(z)|>\frac{\epsilon}{2}>\frac{\epsilon}{4}>|h'(z)|$.

Thus $h+w$ is not injective, and thus $h+w\notin H$. Because, for any $\epsilon>0$, we can find some $h+w$ such that $\|(h+w)-h\|<\epsilon$ and $h+w\notin H$, the set $H$ is not closed in $C^1([a,b],\mathbb{R})$.

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Thanks for an interesting and simple proof. –  Gastón Burrull May 26 '12 at 21:14
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