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I have read often the following.

Let $U\subset \mathbb R^n$ and $V \subset \mathbb R^m$. If $f: U \rightarrow V$ and $g: V \rightarrow \mathbb R^p$ are smooth functions then $g \circ f: U \rightarrow \mathbb R^p$ is also smooth. That should be because of the chain rule.

I have some understanding problems here. Three questions:

  1. Why is it stated $U\subset \mathbb R^n$ and not $U\subseteq \mathbb R^n$ (same for $V$)? Can't $U = \mathbb R^n$?
  2. Why is it explicitly stated that $g$ maps to $\mathbb R^p$ ? Can $g$ not just map to some $G \subseteq \mathbb R^p$?
  3. Isn't it enought hat $f: U \rightarrow V_1$ and $g: V_2 \rightarrow \mathbb R^p$ with $V_1 \subseteq V_2?$ Or is $V_1 = V_2$ really necessary?

Example to 1. and 2.:

Take $f(x) = x$ and $g(x) = x^2$ which are both smooth. Then $(g \circ f)(x) = g(f(x)) = x^3$ is also smooth, so at least in this case it works without problems. Also $g$ maps only to $\mathbb R_{\geq0} \subseteq \mathbb R$. Are there cases where it will not work this way?

Example to 3.:

Take $g(x) = \log(x + 1)$ and $f(x) = x^2$. So $f: U \rightarrow V_1$ and $g : V_2 \rightarrow G$ with $U = \mathbb R$, $V_1 = \mathbb R_{\geq 0}$ and $V_2 = \mathbb R_{>-1}$ and $G = \mathbb R$. So, $V_1 \subset V_2$. In this case, $(g \circ f)(x)$ is smooth but can the fact that $V_1 \subset V_2$ instead of $V_1 = V_2$ prevent this generally?

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  • $\begingroup$ I think you are overthinking the motivation for the notation. $\endgroup$ – zugzug May 18 at 13:02
  • $\begingroup$ @zugzug Maybe, but maybe not. Notation matters, so I want to be sure. $\endgroup$ – user3137490 May 18 at 13:07
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I think you are overthinking the motivation for the notation.

1) People often use $\subset$ and $\subseteq$ interchangeably. 2) $g$ maps to $\mathbb{R}^p$ because it covers situations where it is not injective. Technically, $g$ has some image, but it ends up in $\mathbb{R}^p.$ 3) If $f: U \to V_1$ and $g: V_2 \to \mathbb{R}^p$ with $V_2\subset V_1,$ you would have to be able to extend the domain of $g$ to $V_1$ in order for the composition to be defined. Also, being well-defined doesn't mean smooth.

In your example to 3, $V_1$ is the image of $f$. You wrote it as the domain.

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  • $\begingroup$ 1) I know, but that doesn't answer the question? 2) Ok, but then writing $G \subseteq \mathbb R^p$ should work as well in those cases? Why be more restrictive than neccessary? 3) Right, the example was incorrect, I have updated the question. $\endgroup$ – user3137490 May 18 at 13:53

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