478: Ck-ness of Map from Closed Interval into Subset of Euclidean C∞ Manifold at Boundary Point Equals Existence of One-Sided Derivatives with Continuousness, and Derivatives Are One-Sided Derivatives

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A description/proof of that $C^k$-ness of map from closed interval into subset of Euclidean $C^{\mathrm{\infty }}$ manifold at boundary point equals existence of one-sided derivatives with continuousness, and derivatives are one-sided derivatives

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of map between arbitrary subsets of Euclidean $C^{\mathrm{\infty }}$ manifolds $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.
• The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that $C^k$-ness of any map from any (possibly half) closed interval into any subset of any Euclidean $C^{\mathrm{\infty }}$ manifold at any closed boundary point equals the existence of the one-sided derivatives with continuousness, and the derivatives are the one-sided the derivatives, where $k$ excludes $0$ and $\mathrm{\infty }$.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any Euclidean $C^{\mathrm{\infty }}$ manifolds, ${\mathbb{R}}^d$ and $\mathbb{R}$, any (possibly half) closed interval, $J\subseteq \mathbb{R}$, with the lower and upper boundary points, $t_1<t_2$ ($t_1=t_2$ is excluded because for $[t_1,t_1]$, one-sided derivatives do not make sense), any subset, $S\subseteq {\mathbb{R}}^d$, and any map, $f:J\to S$, for any closed boundary point, $t_j\in J$, $f$ is $C^k$ at $t_j$ if and only if $f$ has the one-side derivatives at $t_j$ and the derivatives maps are continuous over an open neighborhood of $t_j$ on $J$, and the derivatives are the one-sided derivatives.

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2: Proof

Let us suppose that $f$ is $C^k$ at $t_j$.

There are an open neighborhood, $U_{t_j}^{\prime }=(t_j-ϵ,t_j+ϵ)\subseteq \mathbb{R}$, of $t_j$ and a $C^k$ map, $f^{\prime }:U_{t_j}^{\prime }\to {\mathbb{R}}^d$, such that $f^{\prime }{|}_{U_{t_j}^{\prime }\cap J}=f{|}_{U_{t_j}^{\prime }\cap J}$, by the definition.

$d^{k}f/d{t}^k{|}_{t_j}$ is defined to be $d^k{f}^{\prime }/d{t}^k{|}_{t_j}$, but $d^k{f}^{\prime }/d{t}^k{|}_{t_j}$ equals the one-sided derivative of $f$, so, the one-sided derivative exists; besides, $f$ has the derivatives over open $U_{t_j}^{\prime }\cap (J\setminus \{t_j\})$, because $f^{\prime }=f$ there and $f^{\prime }$ has the derivatives there, and as $d^k{f}^{\prime }/d{t}^k:U_{t_j}^{\prime }\cap J\to {\mathbb{R}}^d$ is continuous, $d^{k}f/d{t}^k:U_{t_j}^{\prime }\cap J\to {\mathbb{R}}^d$ (the value at $t_j$ is the one-sided one) is continuous.

Let us suppose that $f$ has the one-side derivatives at $t_j$ and there is an open neighborhood, $U_{t_j}^{\prime }=(t_j-ϵ,t_j+ϵ)\subseteq \mathbb{R}$, of $t_j$ and $d^{k}f/d{t}^k:U_{t_j}^{\prime }\cap J\to {\mathbb{R}}^d$ (the value at $t_j$ is the one-sided one) is continuous.

Let us suppose that $t_j$ is the lower boundary point, $t_j=t_1$. $ϵ$ can be taken to be $t_j+ϵ<t_2$. Denoting the $l$-th one-sided derivative as $a_l$, let us define $f^{\prime }:U_{t_j}^{\prime }\to {\mathbb{R}}^d$ as $f^{\prime }=f$ over $U_{t_j}^{\prime }\cap J$ and $f^{\prime }=f(t_1)+a_1(t-t_j)+1/2a_2(t-t_j{)}^2+...+1/k!a_k(t-t_j{)}^k$ elsewhere. $f^{\prime }$ has the derivatives on the whole $U_{t_j}^{\prime }$, and the derivatives maps are continuous over $U_{t_j}^{\prime }$, because the derivatives equal $d^{l}f/d{t}^l$ over $U_{t_j}^{\prime }\cap J=[t_j,t_j+ϵ)$, continuous over the closed subspace of $U_{t_j}^{\prime }$, and equal $a_l$ over $(U_{t_j}^{\prime }\setminus (U_{t_j}^{\prime }\cap J))\cup \{t_j\}=(t_j-ϵ,t_j]$, continuous over the closed subspace of $U_{t_j}^{\prime }$, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.

Let us suppose that $t_j$ is the upper boundary point, $t_j=t_2$.

$ϵ$ can be taken to be $t_1<t_j-ϵ$.

Denoting the $l$-th one-sided derivative as $a_l$, let us define $f^{\prime }:U_{t_j}^{\prime }\to {\mathbb{R}}^d$ as $f^{\prime }=f$ over $U_{t_j}^{\prime }\cap J$ and $f^{\prime }=f(t_1)+a_1(t-t_j)+1/2a_2(t-t_j{)}^2+...+1/k!a_k(t-t_j{)}^k$ elsewhere. $f^{\prime }$ has the derivatives on the whole $U_{t_j}^{\prime }$, and the derivatives maps are continuous over $U_{t_j}^{\prime }$, because the derivatives equal $d^{l}f/d{t}^l$ over $U_{t_j}^{\prime }\cap J=(t_j-ϵ,t_j]$, continuous over the closed subspace of $U_{t_j}^{\prime }$, and equal $a_l$ over $(U_{t_j}^{\prime }\setminus (U_{t_j}^{\prime }\cap J))\cup \{t_j\}=[t_j,t_j+ϵ)$, continuous over the closed subspace of $U_{t_j}^{\prime }$, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.

So, $f$ is $C^k$ at $t_j$.

Anyway, the derivatives at the boundary point are the one-sided derivatives.

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3: Note

$k=\mathrm{\infty }$ is excluded, because the corresponding infinite series, $f^{\prime }=f(t_1)+a_1(t-t_j)+1/2a_2(t-t_j{)}^2+...$, has not been proved to converge.

But the former direction holds also for the $k=\mathrm{\infty }$ case.

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References

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