The following is a quote from Chapter 4 section "Operations" on page 27 of Analysis I by Amann and Escher.
Discussion:
My understanding of this is that if you have an operation $\circledast$ on $Y$, then you can use that to devise a rule on $Y^X$ that works (informally) as follows: for $f, g \in Y^X$, $f \odot g \mapsto f \circledast g$. This notation seems a bit sloppy (and I am open to suggestions for better notation) but what I'm trying to convey is that $f \odot g$ is the rule that sends $x \mapsto f(x) \circledast g(x)$.
I have looked at the subsequent examples mentioned in the text, and they seem to have to do with the fact that if you have a group $G$, ring $R$, vector space $V$, and a nonempty set $X$, then $G^X$ is a group, $R^X$ is a ring, and $V^X$ is a vector space. (Example 12.11 (a) has to do with algebras and it's currently above my pay grade, so I can't comment on that.)
Questions:
Is my understanding correct? What is the significance of this construction (anywhere, pure or applied)?
The notation "$f \odot g \mapsto f \circledast g$" is meaningless. However, your understanding is correct (I'm just rewording slightly):
And the proper notation is that $f\odot g$ is the rule $x\mapsto f(x)\circledast g(x)$. Or if you want to write it a different way, we can say that given an operation $\circledast:Y\times Y \to Y$, we get an induced operation $\odot:Y^X \times Y^X \to Y^X$.
In slightly less formal (but very common) terminology, we also say that $\odot$ is an operation obtained by applying $\circledast$ pointwise. We use the term "pointwise" for the obvious reason that $f\odot g$ is that function which is defined so that when evaluated at a point $x$, it is simply $f(x)\circledast g(x)$.
I realize at this stage of the text they haven't introduced too many examples yet because they're trying to develop things systematically and logically, but I think it is worthwhile to take small leaps of faith and skip the logic slightly to see specific examples of these constructions.
For now, take $Y = \Bbb{Q}$ and consider the multiplication operation $\Bbb{Q}\times \Bbb{Q}\to \Bbb{Q}$, which we usually denote as a dot $\cdot$ (just use your common-sense elementary school knowledge here). Now, suppose I have two functions $f,g:\Bbb{Q}\to \Bbb{Q}$ (say for example $f(x) = 3x$ and $g(x) = x^2$). If I were to ask you to naively multiply these functions together, you would say that their product is the function $x\mapsto 3x^3$, and the way you would usually express this is: \begin{align} (f\cdot g)(x) = f(x)\cdot g(x) = (3x)\cdot (x^2) = 3x^3 \end{align}
Of course technically the $\cdot$ appearing in $f\cdot g$ is different from the one in $f(x) \cdot g(x)$; this is why they introduce new symbols $\circledast$ and $\odot$ in the text. But of course, these operations are so similar to one another that it is too much of a nuisance to keep inventing new notation, so we just use the same symbol throughout.
Anyway, the point of this construction is that you have an operation on a set $Y$; but you then have another set you're interested in (such as $Y^X$), and the question you ask is "can I get a very similar operation on the new set from the old set". The answer is yes, and the nice thing is that you don't have to come up with a completely arbitrarily defined operation. "Induced" structures are nice because in some sense they're "the most reasonable thing" to do.
Usually, for spaces of functions, such a construction is very useful because it gives you more examples of certain spaces, and maybe sometimes, by studying the new space (which is usually a certain space of functions), you can understand the old space much better. I'm sure as you continue reading and studying more linear algebra/ abstract algebra you'll understand this much better.