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I am reading proof of the chain rule for multivariate functions from Multidimensional Real Analysis:

Let $U \subset R^n$ and $V \subset R^p$ be open and consider the mappings $f: U \rightarrow R^p$ and $g:V \rightarrow R^q$. Let $a \in U$ be such that $f(a) \in V$. Suppose f is differentiable at $a$ and $g$ at $f(a)$. Then $g \circ f : U \rightarrow R^q$ is differentiable at $a$, and we have:

$D(g \circ f)(a) = Dg(f(a)) \circ Df(a) \in Lin(R^n, R^q)$

As part of the proof, we have two operator valued functions:

$\phi: U \rightarrow Lin(R^n, R^p)$ and $\psi: V \rightarrow Lin(R^p, R^q)$ that are continuous at $a$ and $f(a)$ respectively with $lim_{x \rightarrow a} \phi(x) = Df(a)$ and $lim_{y \rightarrow f(a)} \psi(y) = Dg(f(a))$.

The author claims that: $lim_{x \rightarrow a}(\psi(f(x)) \circ \phi(x)) = \psi(f(a)) \circ \phi(a)$. However, I am unable to prove this claim myself.

I would appreciate any help or prod in the right direction. Thank you very much.

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For $y\in{\bf{R}}^{p}$ with $\|y\|\leq 1$, then \begin{align*} &\|\psi(f(x))\circ\phi(x)(y)-\psi(f(a))\circ\phi(a)(y)\|\\ &=\|\psi(f(x))(\phi(x)y)-\psi(f(a))(\phi(a)y)\|\\ &=\|\psi(f(x))(\phi(x)y)-\psi(f(a))(\phi(x)y)+\psi(f(a))(\phi(x)y)-\psi(f(a))(\phi(a)y)\|\\ &\leq\|\psi(f(x))-\psi(f(a))\|\cdot\|\phi(x)y\|+\|\psi(f(a))\|\cdot\|\phi(x)y-\phi(a)y\|\\ &\leq\|\psi(f(x))-\psi(f(a))\|\cdot\|\phi(x)\|\cdot\|y\|+\|\psi(f(a))\|\cdot\|\phi(x)-\phi(a)\|\cdot\|y\|, \end{align*} so \begin{align*} &\|\psi(f(x))\circ\phi(x)-\psi(f(a))\circ\phi(a)\|\\ &\leq\|\psi(f(x))-\psi(f(a))\|\cdot\|\phi(x)\|+\|\psi(f(a))\|\cdot\|\phi(x)-\phi(a)\|. \end{align*} Now $f$ is continuous at $x=a$, so $f(x)\rightarrow f(a)$, and $\psi$ is continuous at $f(a)$, so $\psi(f(x))\rightarrow\psi(f(a))$, so $\|\psi(f(x))-\psi(f(a))\|\rightarrow 0$. Also, $\phi$ is continuous at $x=a$, so $\phi(x)\rightarrow\phi(a)$, so $\|\phi(x)\rightarrow\phi(a)\|\rightarrow 0$. Now note that for a neighbourhood of $x=a$, $\|\phi(x)\|\leq\|\phi(a)\|+1$.

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