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In "Loring W. Tu, An introduction to manifolds" the following question exists: Let $q$ be a point of an $n$-dimensional manifold $M$ and $U$ any neighborhood of $q$. Construct a smooth bump function at $q$ supported in $U$. I answered that question but I want to make sure. Here is my answer:

Let $q$ be arbitrary of $M$ that is contained in a neighborhood $U\subset M$. Then, there exists a coordinate chart $(V,\phi)$ in the maximal atlas of $M$ such that $q\in V \subset U$. In particular, there exists a smooth bump function $\rho:\mathbb{R}^n \to \mathbb{R}$ at $\phi(q)$ supported in $\phi(V)$ that is identically $1$ in a neighborhood $B_r(\phi(q)) \subset \phi(V)$, say, of $\phi(q)$. Define a map $f:M\to \mathbb{R}$ by $$ f(p) = \begin{cases} \rho\bigl(\phi(p) \bigr), &\text{$p\in V$}, \\ 0, &\text{$p\not\in V$}. \end{cases} $$ Being the composite of two smooth function, $f$ is smooth on $V$ and hence on the whole manifold $M$. If $p\in \phi^{-1}\bigl( B_r(\phi(q)) \bigr)$, then $\phi(p) \in B_r\bigl( \phi(q) \bigr)$ and therefore, by the construction of $\rho$, $\rho\bigl( \phi(p) \bigr)=1$. That is, $f \equiv 1$ on the neighborhood $\phi^{-1}\bigl( B_r(\phi(q)) \bigr)$ of $q$. Clearly, by the definition of $f$, $supp\, f \subset V \subset U$. Hence, $f$ is a smooth bump function at $q$ supported in $U$.

Can anyone please revise my proof ?. I appreciate your help. Thanks in advance.

Note: A smooth bump function $f:M\to \mathbb{R}$ at a point $q\in M$ supported in $U\subset M$ is a non-negative smooth function such that $f\equiv 1$ on a neighborhood $V_q \subset U$ of $q$ and that $supp\, f \subset U$.

Hussein Eid
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  • Is $\phi^{-1}(B_r(\phi(q)))$ relatively compact? – Michael L. Jul 06 '17 at 05:53
  • What is meant by the term "relatively compact" ? @MichaelLee – Hussein Eid Jul 06 '17 at 05:55
  • "Relatively compact" = "precompact" = "has compact closure" – Michael L. Jul 06 '17 at 05:56
  • There is nothing in the proof mentioning that $\phi^{-1}(B_r(\phi(q)))$ need necessarily be a relatively compact. @MichaelLee – Hussein Eid Jul 06 '17 at 05:58
  • There is something in the definition of bump function that requires it, if I recall correctly. – Michael L. Jul 06 '17 at 05:59
  • A smooth bump function $f:M\to \mathbb{R}$ at a point $q$ supported in an open set $U\subset M$ is a non-negative smooth function such that $f \equiv 1$ in a neighborhood $V_q \subset U$ and $supp, f \subset U$. @MichaelLee – Hussein Eid Jul 06 '17 at 06:03
  • That's fine, then. I was going off of this definition: http://mathworld.wolfram.com/BumpFunction.html – Michael L. Jul 06 '17 at 06:04
  • You don't stipulate that bump functions be compactly supported? – Michael L. Jul 06 '17 at 06:09
  • I edited the post. Kindly take a look at the "Note". @MichaelLee – Hussein Eid Jul 06 '17 at 06:13
  • I would say that the proof looks mostly good to me, then. I'd probably like some detail as to why $f$ is smooth at the boundary of $V$. – Michael L. Jul 06 '17 at 06:22
  • Because $f$ is identically zero on the boundary of $V$, and any constant function is smooth. @MichaelLee – Hussein Eid Jul 06 '17 at 06:25
  • Being $0$ on the boundary of $V$ does not imply being smooth at the boundary of $V$. Take $f(x) = \sin(x)$ on $(0, \pi)$ and $f(x) = 0$ elsewhere. This is continuous, sure, but not smooth at $x = 0$ or at $x = \pi$. – Michael L. Jul 06 '17 at 06:33
  • Why $f$ is not differentiable at $x=0$ or at $x=\pi$ ?. By the way, $f$ is not even continuous, see the graph @MichaelLee – Hussein Eid Jul 06 '17 at 06:42
  • However the smoothness of $f$ outside $V$ can be proven: If $p \not \in V$, take a coordinate chart $(W,\psi)$ in the maximal atlas of $M$ about $p$, then the composition $f \circ \psi^{-1}: \psi(W) \subset \mathbb{R}^n \to \mathbb{R}$ is indeed the zero map and therefore smooth on $\psi(W)$. In particular, $f\circ \psi^{-1}$ is smooth at $\psi(p)$. By definition, $f$ is smooth at $p$ . – Hussein Eid Jul 06 '17 at 06:47
  • Because in that example, the left-hand and right-hand derivatives do not match at $x = 0$ and $x = \pi$ (left-hand derivative at $x = 0$ is $0$ and right-hand derivative is $1$; left-hand derivative at $x = \pi$ is $-1$ and right-hand derivative is $0$). Typically, we take this to mean that $f$ is not differentiable at that point (see $f(x) = \lvert x\rvert$ for another example of this). – Michael L. Jul 06 '17 at 06:52
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    Smoothness on $V$ and smoothness on $V^c$ are together not equivalent to smoothness on $M$. You also need smoothness at the boundary. – Michael L. Jul 06 '17 at 06:53
  • $M=V \cup V^c$, that is; smoothness on $M$ is equivalent to smoothness on $V$ and $V^c$. – Hussein Eid Jul 06 '17 at 06:56
  • And, yes, my example for $f$ is indeed continuous. Are you sure you graphed the correct function? – Michael L. Jul 06 '17 at 06:57
  • No, smoothness on $M$ is not equivalent to smoothness on $V$ and $V^c$. As an easy counterexample of this, take $g(x) = 1$ on $V = (0, 1)$ and $g(x) = 0$ on $V^c = (-\infty, 0]\cup [1, \infty)$. This is smooth on $V$ and smooth on $V^c$ and is not smooth on $\mathbb{R}$. It's not even continuous on $\mathbb{R}$. – Michael L. Jul 06 '17 at 06:59
  • Your last counterexample is not smooth because it is not continuous. But my function $f$ is continuous. @MichaelLee . This counterexample does not work here.

    Yes, I am sure that the function you gave ($f(x)=\sin x$ on $(0,\pi)$ and $0$ elsewhere) is not continuous. Imagine its graph.

     – Hussein Eid Jul 06 '17 at 07:11
  • Alright, $h(x) = \lvert x\rvert$, then. This is continuous and smooth on $(-\infty, 0]$ and on $(0, \infty)$ and is still not smooth on $\mathbb{R}$. – Michael L. Jul 06 '17 at 07:13
  • And, yes, $g$ is not continuous because it doesn't have to be to be smooth on $V$ and on $V^c$, just continuous on each of those sets. But, it does have to be continuous to be smooth on $\mathbb{R}$, which by itself should tell you that these are not equivalent conditions! – Michael L. Jul 06 '17 at 07:15
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    Also, here's a graph of $f$: http://imgur.com/a/tAfsN. I promise you, it's continuous. – Michael L. Jul 06 '17 at 07:18

1 Answers1

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As pointed out in the comments, the map $f$ that you constructed may not be $C^{\infty}$. But you are almost there.

Notice that in Tu's book the bump function $\phi:\mathbb{R}^{n}\rightarrow \mathbb R$ could be chosen so that its support $\mathrm{supp}\phi$ is a compact subspace of $\mathbb{R}^{n}$. Then $\phi^{-1}(\mathrm{supp}\rho)$, being the image of a compact subspace under the continuous map $\phi^{-1}:\phi(V)\rightarrow V$, is a compact subspace of $V$. Hence $\phi^{-1}(\mathrm{supp}\rho)$ is a compact subspace of $M$. But $M$ is Hausdorff, so this means that $\phi^{-1}(\mathrm{supp}\rho)$ is closed in $M$. Therefore, we have that $$\mathrm{supp}f=\mathrm{cl}_{M}((\rho\circ\phi)^{-1}(\mathbb{R}^\times))=\mathrm{cl}_{M}(\phi^{-1}(\rho^{-1}(\mathbb{R}^\times))\subset\mathrm{cl}_{M}(\phi^{-1}(\mathrm{supp}\rho))=\phi^{-1}(\mathrm{supp}\rho)\subset V.$$ From this follows that $f$ is smooth: If $p\notin V$, choose a chart about $p$ disjoint from $\phi^{-1}(\mathrm{supp}\rho)$. (This is possible because $\phi^{-1}(\mathrm{supp}\rho)$ is closed in $M$.) Then this chart is also disjoint from $\mathrm{supp}f$, so on this chart $f$ is identically $0$. Hence $f$ is $C^{\infty}$ at $p$. This proves that $f$ is $C^{\infty}$ at every point not in $V$. By construction, $f$ is also $C^{\infty}$ at every point in $V$. Hence $f$ is $C^{\infty}$.

Finally you check that $f$ satisfies all the properties required as a bump function at $q$ supported in $U$. This should not be difficult.

Ken
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