# Pathological Function

## Main Question or Discussion Point

What is the meaning of a pathological function?

matt grime
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It entirely depends on the context. Pathological in these contexts means, roughly, 'displaying some very bad behaviour that we can justifiably claim is not representative of the 'average' case'.

HallsofIvy
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Often it refers to a function used as a counter-example for what appears to be an obvious statement. For example, there exist non-linear functions that satisfy f(x+y)= f(x)+ f(y) for all real x,y.

matt grime
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And frequently to one thing that is a counter example to lots of related statements.

arildno
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Gold Member
Dearly Missed
Often it refers to a function used as a counter-example for what appears to be an obvious statement. For example, there exist non-linear functions that satisfy f(x+y)= f(x)+ f(y) for all real x,y.
Being non-linear, I would assume that this means there will exist non-integers a, so that for some x, we have the non-equality:
$$f(ax)\neq{af}(x)$$

Could someone be generous enough to provide an example of such a function?
I'm curious..

mathwonk
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i suppose we want a function f:R--->R which is Q linear but not R linear.

If it were R linear it would be multiplication by a real constant.

To be Q linear it suffices ti define it on a Q basis for R, which exists by the Zorn lemma.

Is that enough?

In this case, pathologicl seems only to mean "surprizing", as the functon is quite elementary and nom more pathological than the Zorn lemma.

Of course one may also regard it as surprizing or pathological that zorns lemma is a consequence of the (non pathological?) axiom of choice.

at first encounter some people may regard the heine borel throem as pathological.

i used to think pathological meant the result made you physically ill, as in algebraic geometry causes cancer.

HallsofIvy
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Being non-linear, I would assume that this means there will exist non-integers a, so that for some x, we have the non-equality:
$$f(ax)\neq{af}(x)$$

Could someone be generous enough to provide an example of such a function?
I'm curious..
Actually, I can't give an example of such a function, but I can tell you how to construct one yourself!

Consider the real numbers as a vector space over the rational numbers. There must exist a basis for such a vector space (there exist a basis for any vector space over any field) but the basis clearly must be uncountable so I can't give you an example. I can, however, assert that 1 and e, for example, are independent and so I can construct a basis containing 1, e, and uncountably many other numbers. Define f(1)= 1, f(e)= 2, f(x)= 0 for x any "basis" number other than 1 or e. Define f(x) for any other x "by linearity". That is, write x as a linear combination a1(1)+ a2(e)+ .... Then f(x)= a1+ 2a2. That, since it is a linear function over the vector space, satisfies f(x+ y)= f(x)+ f(y). It also satisfies f(qx)= qf(x) where q is any rational number but not for q equal to an irrational number. In particular, it is not of the form f(x)= Cx and, so, can be shown not to be continuous.

That is a really "pathological" function! If you attempted to draw its graph, you would have to blacken the entire paper. I don't mean by that that its graph contains every point in the plane. It is, after all, a function and so if crosses any vertical line only once. However, what ever pencil or pen you use to draw the graph has a point with some non-zero radius. What is true is that the graph is "dense in the plane"- every point in the plane is within distance $\delta$ of a point of the graph for any $\delta> 0$.

mathwonk
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notice halls explanation is related closely to the answer i hinted at, only clearer.

I wonder if the existence of such a thing is really dependant on AC.

mathwonk
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i think so. i.e. the so called theorem that every vector space has a bsis is apparently equivalent to AC.

Chris Hillman
Informal definition of "pathological function"

Hi all, matt grime and HallsofIvy already mentioned several points:

It entirely depends on the context. Pathological in these contexts means, roughly, 'displaying some very bad behaviour that we can justifiably claim is not representative of the 'average' case'.
Often it refers to a function used as a counter-example for what appears to be an obvious statement. For example, there exist non-linear functions that satisfy f(x+y)= f(x)+ f(y) for all real x,y.
And frequently to one thing that is a counter example to lots of related statements.
That is,

1. "pathological function" is an informal term, not a technical term; it doesn't have a precise definition (at least, not one which is standard or even widely used); for this reason, this term is best illustrated by listing some examples of functions which have often been called "pathological",

2. pathological functions always exhibit behavior which most would feel is counterintuitive (at least, before seeing a proof!); some classic examples include functions which are:

a. "Lebesgue measurable but not Borel measurable",

b. "continuous but not absolutely continuous",

c. "continuous, differentiable with vanishing derivative almost everywhere, but increasing",

d. "absolutely continuous, strictly increasing, and singular on a positive measure set",

3. many theorems in analysis state that "pathology is the norm"!

In the last bullet, I quibble slightly with what matt grime said: in my experience, particularly in the context of discoveries by nineteenth century analysts, "pathological functions" are always "counterintuitive" but their bad behavior often turns out to be, in some sense, "generic". I immediately add that one of the most important insights from analysis is that there are many things you might reasonably mean by "generic", and these can give quite different characterizations ("Baire category" versus "measure" criteria provide a particularly vivid contrast).

A classic example of a "pathological function" which arises naturally at the intersection of elementary number theory with symbolic dynamics is Minkowski's ? function (that's not a typo, it is written with a question mark), which has the following properties:

1. $?$ is a monotonic transformation of the unit interval,

2. $?$ maps the quadratic irrationals to the set of rationals which (when expressed in lowest form) have denominator not a power of two,

3. the corresponding measure $d ?$ is concentrated on a set of Hausdorff measure $\alpha \sim 0.87$, in the sense that integration over any set of smaller dimension yields zero, while for any set of larger dimension there exist functions which integrate to unity,

4. $d ?$ is an invariant measure for the Farey shift, which is closely related to the construction of simple continued fractions (this provides the connection with "generalized Penrose tilings", aka "Sturmian tilings"),

5. said invariant measure in fact makes the Farey shift isomorphic to the Bernoulli shift on two letters, so this shift has measure-theoretic entropy equal to the maximum possible, the topological entropy $\log 2$; the "equally weighted binary choices" here correspond to downward paths in the Farey tree (the tree which organizes the Farey intervals by inclusion).

See the article by Lagarias in Burr et al., The Unreasonable Effectiveness of Number Theory, AMS, 1992.

Speaking of axiomatics, a book worth examining for the surprising contrast between euclidean and hyperbolic plane is Wagon, The Banach-Tarski Paradox, Cambridge University Press, 1995.

Oh, the theorem (or as some would say, "redefinition") in functional analysis mentioned by mathwonk proves ("by axiomatic fiat", if you like) the existence of Hamel bases.

Last edited:
matt grime
Homework Helper
In the last bullet, I quibble slightly with what matt grime said: in my experience, particularly in the context of discoveries by nineteenth century analysts, "pathological functions" are always "counterintuitive" but their bad behavior often turns out to be, in some sense, "generic".
Quite true,though it is perhaps that those examples didn't realize that what they presumed was generic, or wished to be generic, was actually not generic. However, as you say, it is not a term with any solid meaning. And I did put average in quotes (and didn't use the word generic, which is also not a well defined term!).

arildno
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Gold Member
Dearly Missed
Actually, I can't give an example of such a function, but I can tell you how to construct one yourself!

Consider the real numbers as a vector space over the rational numbers. There must exist a basis for such a vector space (there exist a basis for any vector space over any field) but the basis clearly must be uncountable so I can't give you an example. I can, however, assert that 1 and e, for example, are independent and so I can construct a basis containing 1, e, and uncountably many other numbers. Define f(1)= 1, f(e)= 2, f(x)= 0 for x any "basis" number other than 1 or e. Define f(x) for any other x "by linearity". That is, write x as a linear combination a1(1)+ a2(e)+ .... Then f(x)= a1+ 2a2. That, since it is a linear function over the vector space, satisfies f(x+ y)= f(x)+ f(y). It also satisfies f(qx)= qf(x) where q is any rational number but not for q equal to an irrational number. In particular, it is not of the form f(x)= Cx and, so, can be shown not to be continuous.

That is a really "pathological" function! If you attempted to draw its graph, you would have to blacken the entire paper. I don't mean by that that its graph contains every point in the plane. It is, after all, a function and so if crosses any vertical line only once. However, what ever pencil or pen you use to draw the graph has a point with some non-zero radius. What is true is that the graph is "dense in the plane"- every point in the plane is within distance $\delta$ of a point of the graph for any $\delta> 0$.
Cool! 