Does $\partial g(f(x_{0}) \partial f(x_{0})$ mean the product of $\partial g(f(x_{0}))$ and $\partial f(x_{0})$?

Question

I'm reading about chain rule in the multivariable case:

In my understanding, $\partial g\left(f\left(x_{0}\right)\right) \partial f\left(x_{0}\right)$ does not mean the product of $\partial g\left(f\left(x_{0}\right)\right)$ and $\partial f\left(x_{0}\right)$. Instead, it denotes the composition of $\partial g\left(f\left(x_{0}\right)\right) \in \mathcal L(F,G)$ and $\partial f\left(x_{0}\right) \in \mathcal L(E,F)$, i.e. $\partial g\left(f\left(x_{0}\right)\right) \partial f\left(x_{0}\right) = \partial g\left(f\left(x_{0}\right)\right) \circ \partial f\left(x_{0}\right) \in \mathcal L(E,G)$.

It's written in this way because we usually write $Tv$ rather than $T(v)$ if $T$ is a linear transformation. By this way, it reduces the burden of notation.

My question: Could you please confirm if my understanding is correct?

Answer 1

The product of operators is defined as the composition. Therefore, yes, it does mean the product, and yes, it does mean the composition of the operators.

Think for instance about matrices. The product of matrices is defined to be the composition of the corresponding linear operators.