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I’m reading Lee’s Introduction to Smooth Manifolds. When encountering the uniqueness of integral curves (of smooth vector fields), I wonder that why won’t a bump function or a function like $\begin{cases} 0 & \text{, if } x \le 0 \text{,} \\ \mathrm{e}^{-1/x} &\text{, if } x > 0 \end{cases}$ disagree with it. I also recognize that Norton’s dome challenges the uniqueness of Newtonian mechanics. I inspected the case $\mathrm{e}^{-1/x}$ and Norton’s dome, and found that they don’t admit a smooth vector field.

I think we can directly prove it for general cases, and I formalize it as:

Let $F \colon (-\epsilon, \epsilon) \to \mathbb{R}$ be an smooth increasing function that $F(x) = 0$ for $x \in (-\epsilon, 0]$. Prove that $V$ defined by $V(0) = 0$ and $V(y) = F'(F^{-1}(y))$ for $y > 0$ is not smooth at $y = 0$.

My approach: That is, $V(F(t)) = F'(t)$. We have $$ V'(y) = \frac{\mathop{}\!\mathrm{d} V}{\mathop{}\!\mathrm{d} F} = \frac{\mathop{}\!\mathrm{d} V}{\mathop{}\!\mathrm{d} t} \bigg/ \frac{\mathop{}\!\mathrm{d} F}{\mathop{}\!\mathrm{d} t} = F''(t) / F'(t) \text{.} $$ So it suffices to prove that $F'' / F'$ is unbounded as $t \to 0$. (This is checked for $\mathrm{e}^{-1/x}$ and Norton’s dome.)

Let $f(t) = F'(t)$, we have $f(t) \ge 0$ and the goal becomes $f' / f$ unbounded.

I’m not able to finish the rest of the proof. I tried to use the mean value theorem:

  • For any $x_1 > 0$, we can find an $x_2 \in (0, x_1)$ that $f'(x_2) = f(x_1) / x_1$.
  • Then repeat this we can find $x_n$ based on $x_{n - 1}$.
  • Now consider $f'(x_{n + 1}) / f(x_{n + 1})$, which equals $\displaystyle \frac{1}{x_n} \frac{f(x_n)}{f(x_{n + 1})}$.
  • If we can show that there’s at least one $n$ s.t. $f(x_n) > f(x_{n + 1})$, then by the arbitrariness of $x_1$, we have one $\displaystyle f'(x_{n + 1}) / f(x_{n + 1}) = \frac{1}{x_n} \frac{f(x_n)}{f(x_{n + 1})} > \frac{1}{x_n}$ unbounded.
  • I’m stuck at showing there’s at least one $n$ s.t. $f(x_n) > f(x_{n + 1})$, as we cannot surely state that $x_n \to 0$ as $n \to \infty$.

If anyone can fix the problem in my proof, thanks in advance. Also, I’m looking forward to a simpler proof, too.

Also, I suggest that this question {Why the derivatives $f^{(n)}(x)$ of Flat functions grows so fast? (intuition behind)} may help(?), but I cannot solely make use of it.

Ted Shifrin
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PinkRabbit
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1 Answers1

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It’s very suspicious that @jimmyb posted many answers in a short time, having an obviously wrong answer to my question at first. Isn’t it an LLM-bot?

Nevertheless, thanks to @jimmyb providing a plausible path (in the middle of the answer, the definition of $H(t) = \ln(F'(t))$), although the final argument makes no sense. Now I may proceed:

  • Let $g(t) = \ln(f(t))$, then $g'(t) = f'(t) / f(t)$.
  • It suffices to prove that $g'(t)$ is unbounded.
  • Note that $g(t) \to -\infty$ as $t \to 0$.
  • Assume $g'(t)$ is bounded around $0$, say $\lvert g'(t) \rvert \le M$.
  • We have $g(t) = g(t_0) - \int_{t}^{t_0} g'(u) \mathop{}\!\mathrm{d} u \ge g(t_0) - (t_0 - t) M$.
  • Limiting $t$ to $0$ gives $-\infty = \lim_{t \to 0} g(t) \ge g(t_0) - t_0 M$, a contradiction.
PinkRabbit
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