Why do we use the term "equivalent" with Operators but "equal" with Functions?

Question

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

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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?

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

Answer 1

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:

• Triple bar (Wikipedia)
• Equivalence relation (Wikipedia)
• Equality (mathematics) (Wikipedia)
• Similarity (geometry) (Wikipedia)
• Congruence (Wikipedia)
• Leibniz' Law and that good old riddle

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 :

• Operational calculus
• The Heaviside operational calculus
• The Theory Of Linear Operators
• Heaviside’s Operator Calculus
• What is operator calculus?
• Operator (physics)
• Operator methods in quantum mechanics
• Differential equations and Fourier and Laplace transforms
• Why exp(-st) in the Laplace Transform?
• Finite difference
• Exponential of a function times derivative
• Inhomogeneous 2nd-order linear differential equation
• Fractional calculus
• Differential operators confusion
• How to make sense of this calculus notation, Advanced College Level

I suppose that the mathematics for physics and engineering is not different from the mathematics for economics and psychology.

Answer 2

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.