Let f:[a,b] -> R be a real-valued (weakly) monotone function. By
monotone function, we mean one that is always (weakly) increasing:
x < y implies f(x) <= f(y)
or always (weakly) decreasing:
x < y implies f(x) >= f(y)
In either case f will be bounded:
min{f(a),f(b)} <= f(x) <= max{f(a),f(b)}
but needed not be continuous.
[Such functions are, however, of bounded variation and can have at
most a countable number of discontinuities. In proving that at each
point x for which it makes sense, both a left- and right-sided limit
exist, we are in fact showing that any discontinuity must be a
so-called "jump" discontinuity.]
First consider the case that f is increasing and take point x
belonging to (a,b], so that it makes sense to ask about a left-sided
limit of f at x (or the limit of f as we approach x "from below").
Does such a limit exist?
Intuitively the left-sided limit of f at x should be the least upper
bound of f applied to [a,x). To make this rigorous we note that
f[a,x) has a least upper bound s by the least upper bound property of
the real numbers:
[Least Upper Bound Property of the Real Numbers]
http://www.shu.edu/projects/reals/infinity/proofs/lubprop.html
because f(x) is an upper bound of f[a,x) (use the fact that f is
increasing) and f[a,x) is nonempty (it contains f(a)).
It remains to show that f(z) approaches s as z approaches x from below
(the left-sided limit). Let epsilon > 0 be given. According to the
definition of limits, we want to show the existence of delta > 0 such
that:
x - delta <= z < x implies |f(z) - s| < epsilon
To prove the existence of such delta, it suffices to find y in [a,x)
such that:
s - epsilon < f(y)
If no such y were to exist, then s - epsilon would be an upper bound
on f[a,x), contradicting the choice of s as least upper bound. So
pick some y and let:
delta = x - y
Now, if x - delta <= z < x, then by our choice of delta:
y <= z
whence, by the monotonic increasingness of f:
f(y) <= f(z)
Combining this with our choice of y s.t. s - epsilon < f(y), we have:
s - epsilon < f(y) <= f(z) <= s
Thus since |f(z) - s| = s - f(z) and:
s - epsilon < f(z) implies s - f(z) < epsilon
we have:
|f(z) - s| < epsilon
as desired.
Having shown the existence of the left-sided limit for monotone
increasing function f, it should be seen that the remaining cases can
be reduced to this. For if f were instead a decreasing function
(rather than increasing), the prior arguments would apply to -f, and
the existence of left-side limit s for -f at point x is equivalent to
the existence of left-sided limit -s for f at point x.
Furthermore the questions of existence of right-sided limits can be
resolved by applying the results already obtained to function
g:[-b,-a] -> R defined by:
g(x) = f(-x)
For it will be observed that g is monotone increasing (decreasing)
whenever f is monotone decreasing (resp. increasing), and the
existence of a right-sided limit for f at point x is equivalent to the
existence of a left-sided limit for g at point -x.
regards, mathtalk-ga |
