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I'm trying understand the article "Curvature bound for curve shortening flow via distance comparison and a direct proof of Grayson's theorem" by Ben Andrews and Paul Bryan and they stated that

Define $a(p, q) = \inf \ \{e^t : d(p, q) ≥ f(ℓ(p, q), −t) \}$ for $p \neq q$ in $S^1$. Then by the implicit function theorem $a$ is continuous, and smooth and positive where $0 < d < 2 \sin(ℓ/2)$ where it is defined by the identity

$$d(p, q) = f(ℓ(p, q), − log(a(p, q))).$$

I'm stuck in understanding why the Implicit Function Theorem ensures that $a$ is smooth and why the identity $d(p, q) = f(ℓ(p, q), − log(a(p, q)))$ is valid for $0 < d < 2 \sin(ℓ/2)$? I think that the Implicit Function Theorem is applied on the $C^k$ function

$$F: S^1 \times S^1 \times \mathbb{R}\ - \left( \{ (p,q) \in S^1 \times S^1 \ ; \ p = q \} \times \mathbb{R} \right) \longrightarrow \mathbb{R}$$

defined by $F(p,q,t) := d(p, q) - f(ℓ(p, q), −t)$, because $\frac{\partial F}{\partial t} = \frac{\partial f}{\partial t} > 0$ as computed in the article and exist $(p_0, q_0, t_0) \in \text{Dom} \ F$ such that $F(p_0,q_0,t_0) = 0$ (we can ensure this because we choose $t_0$ so that $d(p_0, q_0) = f(ℓ(p_0, q_0), −t_0)$), then exist open sets $U$ containing $(p,q)$ and $V$ containing $t$ so that $g: U \longrightarrow V$ is a function $C^k(U)$ and $F\left(p, \ q, \ g(p,q)\right) = 0$ for all $p,q \in U$.

The Implicit Function Theorem ensure that exist a function $g$, but the theorem doesn't ensure that the function $a$ is the function $g$. I didn't understand why the smoothness of $a$ is ensured by Implicit Function Theorem. If anyone can help me in understand this and why $d(p, q) = f(ℓ(p, q), − log(a(p, q)))$ for $0 < d < 2 \sin(ℓ/2)$, I will be grateful!

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  • $\begingroup$ Uniqueness (i.e. the correspondence between $a$ and $g$; I think it should actually be $a=e^g$) should follow from the monotonicity of $f$ in $t$. For smoothness, just use the $C^\infty$ version of the implicit function theorem. $\endgroup$ – Anthony Carapetis Apr 9 '18 at 10:34
  • $\begingroup$ @AnthonyCarapetis comment and your reply seems to have resolved the problem. Let me just add that the function $f$ has the same first order term as $d$ and we're effectively saying here that choosing $\bar{t}$ negative enough ensures the next order term of $f$ is smaller than the next order term of $d$ so we get initial inequality near $x=0$. Away from $x=0$ is no problem since $d$ is bounded away from $0$ and $f \to 0$ as $\bar{t} \to 0$. $\endgroup$ – Paul Bryan Apr 11 '18 at 0:41
  • $\begingroup$ @AnthonyCarapetis, I understood now, thanks a lot! Comment your answer please so I can vote. $\endgroup$ – George Apr 12 '18 at 2:45
  • $\begingroup$ @PaulBryan, what did you want to say exactly with "$f$ has the same first order term as $d$"? $\endgroup$ – George Apr 13 '18 at 13:43
  • $\begingroup$ We are trying to get initial inequality and have something like $d = \ell + C\ell^3 + \cdots$ and $f(\ell) = \ell + B(\bar{t})\ell^3 + \cdot$. Then if $B(\bar{t}) < C$, we have that $f < d$ for small lengths $\ell$ - say on $[0, \ell_0)$. We get this since $B(\bar{t}) \to -\infty$ as $t \to -\infty$. To ensure that $f < d$ on $[\ell_0, \pi]$, just note that $f \to 0$ uniformly on $[\ell_0, \pi]$ as $\bar{t} \to -\infty$. It’s just another perspective on how we get initial inequality. $\endgroup$ – Paul Bryan Apr 14 '18 at 22:31

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