Making a limit substitution rigorous

Question

Suppose I'm given the following problem

Let $A \subseteq \mathbb{R}^m$ and let $f : A \to \mathbb{R}^n$. Let $a \in A$ and $u \in \mathbb{R}^n$ such that $u \neq 0$. Show that if $f_u'(a)$ exists, then $f_{cu}'(a)$ exists and $f'_{cu}(a) = c \cdot f_{u}'(a)$

I can prove the above as follows

Proof: Since $f_u'(a)$ exists then by definition $$\lim_{t \to 0} \frac{f(a+ut)-f(a)}{t}$$ exists and equals $f_u'(a)$. Then we have

\begin{align*} \lim_{t \to 0} \frac{f(a+tcu)-f(a)}{t} &= \lim_{t \to 0} \frac{cf(a+tcu)-cf(a)}{ct} \\ &= \lim_{t \to 0} \frac{c(f(a+tcu)-f(a))}{tc} \\ &= \lim_{s \to 0 } \frac{c(f(a+su)-f(a))}{s} \ \ \ \ \ \ \text{ where $s = tc$} \end{align*}

and now comparing to the fact that $\lim_{t \to 0} \frac{f(a+ut)-f(a)}{t}$ exists and equals $f'_u(a)$ it follows (using another limit theorem) that $$\lim_{s \to 0 } \frac{c(f(a+su)-f(a))}{s} = c \cdot \left(\lim_{s \to 0 } \frac{f(a+su)-f(a)}{s}\right) = c \cdot f_u'(a)$$ hence $f'_{cu}(a) = c \cdot f_{u}'(a)$ $\square$

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But now above there's a point where I'd have to make the substitution $s = tc$, and I'd like to make my argument more formal based on the answer given here : https://math.stackexchange.com/a/167948/266135

This is my attempt to do that. If I let $\phi : \Theta \to \mathbb{R}^n$ (where $\Theta = \{ t \in \mathbb{R} \setminus \{0\} \ | \ a+tcu \in A\}$) be defined by $$\phi(t) = \frac{c(f(a+tcu)-f(a))}{tc}$$ and $g :\Gamma \to \mathbb{R}$ (where $\Gamma = \{tc \in \mathbb{R} \ | a+tcu \in A\}$) be defined by $g(tc) = tc$. Then $$\lim_{tc \to 0}g(tc) = 0.$$

Now I want to arrive at $$\lim_{tc \to 0}\phi(g(tc)) = \lim_{t \to 0}\phi(t)$$ but for that I need $\phi$ to be continuous at $\lim_{tc \to 0}g(tc) = 0$, however $\phi$ isn't even defined at $0$.

So my question is this, how could make my substitution argument in my proof above more formal?

Answer 1

I don't understand what all the fuss is here.

Fix $c\ne 0$ and let $g$ be any function defined in a punctured neighborhood of $0$ for which we have $\lim\limits_{s\to 0} g(s)=L$. Then have $\lim\limits_{t\to 0} g(ct) = \lim\limits_{s\to 0} g(s)=L$. You can check this easily from the $\delta$-$\epsilon$ definition. If $|g(s)-L|<\epsilon$ when $0<|s|<\delta$, then $|g(ct)-L|<\epsilon$ when $0<|t|<\delta/|c|$.

Answer 2

Let us say that $f(x) \to L^{\ne}$ as $x \to x_0^{\ne}$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that whenever $0 < d(x, x_0) < \delta$, we have $0 < d(f(x), L) < \epsilon$. This is similar to the usual limit definition, except that we added the requirement $0 < d(f(x), L)$. Similarly, let us say $f(x) \to L$ as $x \to x_0^{\ne}$ if the usual limit $\lim_{x \to x_0} f(x) = L$ holds. Then it it easy to verify the following statements:

• If $f(x) \to L^{\ne}$ as $x \to x_0^{\ne}$ and $g(y) \to M$ as $y \to L^{\ne}$, then $g(f(x)) \to M$ as $x \to L^{\ne}$.
• $f(x) \to L^{\ne}$ as $x \to x_0^{\ne}$ if and only if $f(x) \to L$ as $x \to x_0^{\ne}$ and there is some $\delta > 0$ such that whenever $0 < d(x, x_0) < \delta$, we have $f(x) \ne L$.

Now, if $c \ne 0$, then the function $\phi(t) := ct$ is easily verified to satisfy $\phi(t) \to 0^{\ne}$ as $t \to 0^{\ne}$. Therefore, the composition principle in the first bullet above will apply in your case.

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The language of filters will provide a nice way to generalize the composition principle. In particular, suppose we have some function $f : X \to Y$, and filters $\mathscr{F}$ on $X$ and $\mathscr{G}$ on $Y$, respectively. Then we will say $f(x) \to \mathscr{G}$ as $x \to \mathscr{F}$ if $\mathscr{G} \subseteq f_* \mathscr{F} \overset{def}{=} \{ S \subseteq Y \mid f^{-1}(S) \in \mathscr{F} \}$. It is then easy to prove:

Suppose we have $f : X \to Y$, $g : Y \to Z$, and filters $\mathscr{F}$, $\mathscr{G}$, and $\mathscr{H}$ on $X, Y, Z$ respectively. Suppose, furthermore, that $f(x) \to \mathscr{G}$ as $x \to \mathscr{F}$ and that $g(y) \to \mathscr{H}$ as $y \to \mathscr{G}$. Then $g(f(x)) \to \mathscr{H}$ as $x \to \mathscr{F}$.

Also, when we have filter bases, it is often useful to use this criterion (which is easy to prove):

Suppose we have a basis $\mathcal{B}$ for the filter $\mathscr{G}$. Then $f(x) \to \mathscr{G}$ as $x \to \mathscr{F}$ if and only if for each $N \in \mathcal{B}$, we have $f^{-1}(N) \in \mathscr{F}$.

Some common examples of filters which would be useful in combination with this notation:

• $a^{\ne}$ as notation for the filter of punctured neighborhoods near $a$ (which has basis being the punctured balls $B_\epsilon '(a)$ for $\epsilon > 0$).
• $a$ as notation for the filter of neighborhoods of $a$ (which has basis given by the open balls $B_\epsilon(a)$ for $\epsilon > 0$).
• For $a \in \mathbb{R}$, $a^+$ as notation for the filter generated by $(a, a + \epsilon)$ for $\epsilon > 0$.
• Similarly, for $a \in \mathbb{R}$, $a^-$ as notation for the filter generated by $(a - \epsilon, a)$ for $\epsilon > 0$.
• $\infty$ as notation for the filter on $R$ generated by $[M, \infty) \subseteq \mathbb{R}$ for $M \in \mathbb{R}$; and similarly, $-\infty$ as notation for the filter generated by $(-\infty, M]$ for $M \in \mathbb{R}$; and also $\pm \infty$ as notation for the filter generated by $(-\infty, M] \cup [M, \infty)$ for $M \in \mathbb{R}$.
• Similarly, for use in working with limits of sequences: $\infty$ as notation for the filter on $\mathbb{N}$ generated by $\{ n \in \mathbb{N} \mid n \ge N \}$ for $N \in \mathbb{N}$.
• A filter on $\mathbb{C}$ with notation $|z| \to \infty$ which is the filter generated by $\{ z \in \mathbb{C} \mid |z| \ge R \}$ for $R \in \mathbb{R}$. (And similarly, on any normed vector space $X$, we have $\lVert x \rVert \to \infty$ as notation for the filter generated by $\{ x \in X \mid \lVert x \rVert \ge R \}$ for $R \in \mathbb{R}$.)

We can also prove more convenient criteria for limits with respect to some of these filters. For example:

Suppose we have $f : X \to \mathbb{R}$, $L \in \mathbb{R}$, and $\mathscr{F}$ a filter on $X$. Then $f(x) \to L^+$ as $x \to \mathscr{F}$ if and only if $f(x) \to L$ and for some $N \in \mathscr{F}$, $f(x) > L$ for all $x \in N$.

So, from this point of view, it's a bit unfortunate in my opinion that the usual notation $\lim_{x \to x_0} f(x) = L$ looks like it's more or less symmetric with respect to domain and codomain, whereas it's actually using two different types of filters on the two spaces: it is equivalent in our notation to $f(x) \to L$ as $x \to x_0^{\ne}$.

(Note that this is mostly notation that I've developed for myself in thinking about these types of limit composition issues. If there's more standard notation for this concept, feel free to let me know and/or edit it in.)