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Are the following true?

  • $\operatorname{scalar} \circ \operatorname{function} = \operatorname{scalar} \times \operatorname{function}$

  • $\operatorname{function} \circ \operatorname{scalar} = $ the function evaluated at that scalar.

Background:

Let $f:E \to F, g: F \to G$, where $E,F$ and $G$ are normed vector spaces. The chain rule tells us: if $f$ is differentiable at $x_0$ and $g$ is differentiable at $f(x_0)$, then $g \circ f$ is differentiable at $x_0$, and:

$$D(g\circ f)(x_0) = Dg(f(x_0)) \circ Df(x_0)$$

I'm having some confusions when it happens that one of these mappings is a scalar.

For instance, to find the differential of the square root of the inner product on a real pre-Hilbert space ($0$ excluded), we would define the mapping $x \to \langle x, x \rangle$, find its differential, then consider the square root map on $(0, \infty)$. The square root of the inner product is just the composition of these two. But how does its differential map look like?

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  • $\begingroup$ Your formula of the chain rule is wrong! The correct formula is $$D(g \circ f) (x_0) = Dg(f(x_0)) \cdot Df(x_0)$$ where $\cdot$ denotes multiplication. $\endgroup$ Commented Oct 29, 2015 at 23:50
  • $\begingroup$ Not really @Crostul. Anyway, how would you define the multiplication of a function which takes as arguments elements in $F$, and one which takes as arguments elements of $E$? $\endgroup$ Commented Oct 29, 2015 at 23:52
  • $\begingroup$ Both are correct. $Dg$ and $Df$ are linear maps. Multiplication of linear maps is defined as their composition. $\endgroup$ Commented Oct 30, 2015 at 0:23

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If $V$ is your pre-Hilbert space (which I am assuming is real), then let $$f:\ V \to \Bbb R\ :\ x \mapsto \langle x, x \rangle$$ Then for each $x \in V, Df|_x$ is a linear map from $V \to \Bbb R\ :\ v \mapsto 2\langle x, v\rangle$. Let $$g\ :\ \Bbb R \to \Bbb R\ :\ t \mapsto \sqrt t$$ Then $Dg|_t$ is a linear map from $\Bbb R \to \Bbb R\ :\ q \mapsto \frac{q}{2\sqrt t}$.

Hence $$\begin{align} D(g\circ f)|_x(v)&= Dg|_{f(x)}\circ Df|_x(v) \\&=Dg|_{\langle x, x\rangle}(2\langle x, v\rangle) \\&=\frac{2\langle x, v\rangle}{2\sqrt{\langle x, x\rangle}} \\&=\frac{\langle x, v\rangle}{\|x\|}\end{align}$$

You have to remember that the differential is always a linear map. Even when we choose to represent it as a scalar, this is only because linear maps $\Bbb R \to \Bbb R$ are just scalar multiplication.

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