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Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the graph of the absolute value function. That is, $X=\{(x,|x|) : x\in\mathbb{R})\}$.

We define a functional structure on $X$ by restricting smooth real valued functions defined on open subsets of the plane: an open set $U\subset X$ has the form $U=V\cap X$, where $V\subset\mathbb{R}^{2}$ is open. Then define $f\in F(U)$ iff $f=g|_{U}$, where $g$ is a smooth real value function on $V$.

E.g., The function $\pi_{2}(x,y)=y$ is smooth on any neighborhood $V$ of the origin. Therefore, its restriction to $U=V\cap X$ is in $F(U)$ (note that the restriction is not smooth at the origin).

Question: Is $\mathbf{X}$ with the functional structure $\mathbf{F}$ a differentiable manifold? (Defining smooth manifolds via functional structures: here).

It seems to me that it is not a differentiable manifold. I tried showing that at the origin it is not locally isomorphic to $(\mathbb{R},C^{\infty})$ (the real line with the functional structure of smooth real valued functions). Assuming that there is a local isomorphism at the origin I was trying to get to a contradiction more or less using functions such as $\pi_{2}$ above, which are not smooth when restricted to $X$. Nothing has worked so far...

Note: Clearly $X$ is not a manifold in $\mathbb{R}^{2}$. This is discussed for example in this and this posts.

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The smooth structure on $\mathbb R$ has the following property: if $f\in C^\infty(\mathbb R)$ and $f\equiv 0$ on some half-line (a connected component of $\mathbb R\setminus \{a\}$), then there is $h\in C^\infty(\mathbb R)$ such that $h(a)=0$ and $$\lim_{x\to a}\frac{f(x)}{h(x)}=0 \tag1$$ The above basically says that $f'(a)=0$, but is stated solely in terms of the functional structure. Consequently, the property is preserved by isomorphisms.

However, $(X,F)$ does not have the above property. Indeed, the smooth function $g(x,y)=x+y$ restricts to a function $f$ on $X$ that vanishes on a connected component of $X\setminus \{(0,0)\}$. There is no smooth function $h$ on $X$ for which (1) holds, because such a function would have to be non-Lipschitz at $(0,0)$.

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I think I follow, but just in case, could you clarify what you mean when saying "Lipschitz at $(0,0)$". –  John Aug 27 '13 at 20:27

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