A (real valued) function is of bounded variation if, roughly
speaking, it doesn't "wiggle" too much. Let's make that definition
Let a=x0<x1<...<xn=b be a
partition of a closed bounded interval. We'll call this partition
P. We total the variation of f as we move from one point of P to
the next. That is, let
V(f,P)=sum|f(xi)-f(xi-1)| (i goes from 1 to n).
Now it's not hard to prove (eg by induction) that if P⊂Q (that is,
Q is a "finer partition" than P), then V(f,P)≤V(f,Q). The question we
ask ourselves is whether V(f,P) can be made as large as we like, as the
partition P gets finer and finer. If not, then we say f is of bounded
variation (on the interval in question). That is, f is of bounded
V(f)=sup V(f,P)<∞, in which case we call V(f) the variation of f.
Now for a few simple facts about functions of bounded variation.
I'll leave the reader to verify them. First, such functions are
bounded. Also, all monotonic (increasing or decreasing) functions
are of bounded variation (in fact, V(f)=|f(b)-f(a)|, while if f is not
monotonic, V(f)>|f(b)-f(a)|). Finally if c is a real constant, and
f, g are of bounded variation, then cf and f+g are of bounded variation
(with V(cf)=|c|V(f) and V(f+g)≤V(f)+V(g)). That is, the set of
functions of bounded variation forms a vector space over the reals.
In particular, any function which can be written as the difference of
two increasing functions is of bounded variation.
Now it can be shown that if f is of bounded variation on the
closed interval a,b, and a≤x≤b, then f is of bounded variation (in
fact, the variation from a to x, plus the variation from x to b, equals
the variation from a to b, which should remind you of the definite
integral). Consequently, the variation of f from a up to x defines a
function v(x). Clearly v is increasing, and using the result above
that |f(b)-f(a)|≤V(f), we can now show that u(x)=v(x)-f(x) is also
increasing. That is if f is of bounded variation, it can be written as
the difference of two increasing functions.
Comparing the remarks at the end of the 2 previous
paragraphs, we have that f is of bounded variation iff it can be
written as the difference of 2 increasing functions. This is known as
Some simple examples of functions which are not of bounded variation
are the characteristic function of the rationals,
or, if you prefer a continuous example, f(x)=x sin(1/x) for x non
zero, and 0 for x=0 (on any interval containing 0).
It can be shown that f is continuous at a point iff v (and hence
also u), as defined above, is. Remark: if f is differentiable in the
interval, with bounded derivative, then f is of bounded variation (this
follows from the mean value theorem).
Update: User maxClimb
suggested I also mention why anyone would care about such functions. The first thing I would say is that I'm a pure mathematician
, so we don't ask such questions
... However, there are reasons. On the pure side, such functions are especially "nice" - they're continuous and differentiable almost everywhere
, and they're precisely those functions with respect to which one can perform Riemann-Stieltjes integration
on continuous functions. On the more applied side, they are essentially the only type of functions which appear in most physical systems (because of their niceness), and so an understanding of their properties is invaluable to understanding said physical system.