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Why do we speak in terms of "equality" when we deal with functions but "equivalence" when dealing with operators?

To elaborate:

Two functions, f and g are equal to each other (denoted: f=g) if:

  1. They share the same domain.
  2. For every x in this domain, the values of f and g evaluated at x are equal.

In contrast, "two operators, O₁ and O₂ are said to be equivalent (denoted: O₁≣O₂) if for any function y to which O₁ and O₂ are each applicable, the functions O₁y and O₂y are equal."[1]

Could someone please give me the rationale behind this distinction? Does this imply that operators are only equivalence relations while functions are equivalence relations and partial orders? And, if so, why are operators not also partial orders?

[1] Quoted from Samuel Goldberg's Introduction to Difference Equations.

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Steve S
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  • @GitGud: I like your answer. It might not be right but still I like it. – Steve S Jul 30 '14 at 19:48
  • I removed my comment because I had read it as 'differential equations'. – Git Gud Jul 30 '14 at 19:49
  • Not in the (quantum mechanics) textbooks I've been reading (quite a while ago). There they simply wrote $O_1 = O_2$ . – Han de Bruijn Jul 30 '14 at 20:01
  • @HandeBruijn: Really?? Because at this point--after spending so much time composing my original question--I really want there to be some enlightening answer to come out of this (not merely "personal preference" or something like that...). – Steve S Jul 30 '14 at 20:33
  • The wording of the quote seems to indicate that the domains to which $O_1$ and $O_2$ apply could be different but have some intersection. – Hagen Knaf Jul 31 '14 at 07:11

2 Answers2

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Let us denote the domain of a function (or operator) $f$ by $\mathcal D(f)$. Then, by your definition, $f=g$ if and only if $\mathcal D(f)=\mathcal D(g)$, and $f(x)=g(x)$ for all $x\in\mathcal D(f)$. On the other hand, $O_1$ and $O_2$ are equivalent if and only if $O_1(f)=O_2(f)$ for all $f\in \mathcal D(O_1)\cap\mathcal D(O_2)$. So equality is a stronger notion than equivalence.

Jonas Dahlbæk
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    This is a good point. The question remains: was this just a mistaken omission, or was it intentional? – goblin GONE Aug 01 '14 at 11:19
  • In any case, I have never seen this use of the word equivalent before. – Jonas Dahlbæk Aug 01 '14 at 12:21
  • Yes, I think it is poorly chosen terminology, since this relation is not transitive. Its still an interesting relation though; partial functions $f$ and $g$ are equivalent iff $f \cup g$ is a partial function. That is interesting. – goblin GONE Aug 01 '14 at 12:23
  • I must admit I find the notion a bit lacking in the sense that any two operators whose domain have trivial intersection are 'equivalent'. Perhaps it would be more satisfying to have some type of condition on the intersection such as it being a dense set or a core for each operator. – Jonas Dahlbæk Aug 01 '14 at 12:27
  • But that's kind of the point; if they have empty intersection, then the partial functions cannot possible disagree, so we can safely take their union. – goblin GONE Aug 01 '14 at 12:29
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Equal ($=$), equivalent ($\equiv$), identical ($\equiv$), congruent ($\cong$), similar ($\sim$), if and only if ($\Longleftrightarrow$).
Can anyone tell what exactly is the difference between these things ?
Some material to think about:

In short, I don't think that the book by Samuel Goldberg, with "applications in the social sciences, economics, and psychology", is enough of a reference to underpin your judgment that there would be a consent about speaking in terms of "equality" when we deal with functions but "equivalence" when dealing with operators. As far as I can see, there is no such consent and the choice is typical for the book at hand. Given the similar (!) properties of $=$ and $\equiv$ , it doesn't matter much either.

EDIT. Quoting references as requested. Everywhere $=$ is used instead of $\equiv$ for operator identity :

I suppose that the mathematics for physics and engineering is not different from the mathematics for economics and psychology.
Han de Bruijn
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  • Then quote a reference (bearing sufficient mathematical gravitas, of course) that refutes Goldberg--don't just cop a judgement based on a bias against the intended audience. Don't get me wrong--it's entirely possible that your conclusion is correct. It's just that your rationale seems, well, a little unfair. – Steve S Jul 31 '14 at 20:35
  • You're citing things that you've written yourself?! [I bet you didn't think I'd actually check... :) ]. I guess I was hoping for a reference from a book. You know, something with more gravitas... – Steve S Aug 04 '14 at 05:58
  • Yes indeed, I've written myself less than seven percent of the references. That is good when compared with the fact that your notation $O_1 \equiv O_2$ comes from just one reference [1]. And people here are volunteers. So how about going to a library and digging up a few more books about Operator Calculus yourself ? – Han de Bruijn Aug 04 '14 at 10:12