Main Question or Discussion Point
What is the meaning of a pathological function?
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: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.
Actually, I can't give an example of such a function, but I can tell you how to construct one yourself!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:
Could someone be generous enough to provide an example of such a function?
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.
That is,And frequently to one thing that is a counter example to lots of related statements.
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!).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".
Cool!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 [itex]\delta[/itex] of a point of the graph for any [itex]\delta> 0[/itex].