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I am working on a problem and I came to a point where the problem has to be considered within a class of functions. But this class cannot be defined just in terms of the properties of each individual function in this class. It has to be defined in terms of the properties of arbitrary two functions in this class.

Let me give you an example (below $m$ is known fixed integer). Let $\mathcal{F}$ be the class of functions defined on $[0,1]$ with the following properties:

  1. Any $F \in \mathcal{F}$ is non-negative and is (weakly) increasing on $[0,1]$.

  2. Any $F \in \mathcal{F}$ is differentiable on $[0,1]$. Moreover, $F'$ is absolutely continuous on $[0,1]$.

  3. $F'(1) > 0$ for any $F \in \mathcal{F}$.

  4. For any $F_1, F_2 \in \mathcal{F}$, if $F_1$ and $F_2$ are not identical, then for the derivatives of these functions it holds that either a) $F'_1(s) (F'_1(1))^{\frac{1}{m}}> F'_2(s) (F'_2(1))^{\frac{1}{m}}$ in a small right side neighbourhood of $0$, or b) $F'_1(s) (F'_1(1))^{\frac{1}{m}}< F'_2(s) (F'_2(1))^{\frac{1}{m}}$ in a small right side neighbourhood of $0$.

  5. For any $F_1, F_2 \in \mathcal{F}$, it holds that if $F'_1(s) (F'_1(1))^{\frac{1}{m}}> F'_2(s) (F'_2(1))^{\frac{1}{m}}$ in a small right side neighbourhood of $0$, then $\frac{F''_1(s)}{F'_1(1)} < \frac{F''_2(s)}{F'_2(1)}$ a.e. in a small left side neighborhood of 1.

My question is: How unusual is it to define sets or classes by describing the properties of any two components in this set rather than just describing the properties of each individual component in the set. Maybe somebody knows examples from other fields where such definitions can be found?

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To answer your question: in the several decades that I've been reading math papers, I haven't ever seen definitions like this that I can recall. My reading's mostly been in topology, however, so I can't speak to other areas of mathematics.

But there's probably a good reason I haven't seen it:

When you define a class this way, it may well be empty, or there may be two mutually exclusive solution sets.

Let me give an example: Suppose I define $G$ to be "the class of functions from the reals to the nonzero reals with the property that if $f$ and $g$ are in the class, then there's a positive real number $c$ with $f = cg$."

Now that's a silly definition, but consider

(a) the set of all constant functions from $\mathbb R$ to $\mathbb R^{+}$, and

(b) the set of all positive multiples of $f(x) = e^x$.

These classes each satisfy my definition of $G$, but are disjoint.

So: you have a job ahead of you. You need to prove that your characterization in fact describes only one possible class, and (probably) that it's nonempty.

Just using the word "the", as in "the class of functions such that..." is already assuming (or at least suggesting very strongly) that the definition is meant to characterize something unique, and this at the very least requires a proof. If you're thinking to yourself "Most definitions I see aren't usually followed by proofs," you're right. There's a reason for that -- it tends to be a bad expository style.

A better approach might be to say that a class has "Property J" if the following 5 things hold. You can then prove statements like "this class has property J" or "that class does not have property J", and the proofs are then associated to theorems rather than to definitions, and your reader will understand, from the statement of the theorem, all the things that need to be proved (and the possibility that there might be multiple sets with property J).

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  • $\begingroup$ Thanks! The problem I am considering is quite difficult and I can reach some conclusions only in a class of functions described above. My class is definitely not empty but it is true that there are many mutually exclusive classes of functions that satisfy the described conditions. In fact, the set of all functions I can potentially be interested in can be split into mutually exclusive classes that satisfy described conditions. $\endgroup$ – MerylStreep Sep 14 '15 at 18:06
  • $\begingroup$ I think there is some flavour of pre-order to my definition of class but not sure how to make use of it. $\endgroup$ – MerylStreep Sep 14 '15 at 18:07
  • $\begingroup$ Well...I'm not sure either. In general, I think that "definitions" should uniquely characterize things. We don't talk about "the span of these three vectors" and mean two different things, after all. When you write "Let F be the class of functions ..." you're implicitly already misleading your reader. $\endgroup$ – John Hughes Sep 14 '15 at 18:35
  • $\begingroup$ I agree that it is better to describe it as a property rather than a definition, and also use "a class", not "the class". $\endgroup$ – MerylStreep Sep 17 '15 at 8:13

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