Increasing and differentiable implies nonnegative derivative that is not identically zero on any interval
Contents
Statement
On an open interval
Suppose is a function on an open interval
that may be infinite in one or both directions (i..e,
is of the form
,
,
, or
). Suppose the derivative of
exists everywhere on
. Suppose further that
is an increasing function on
, i.e.:
Then, for all
. Further, there is no sub-interval of
such that
for all
in the sub-interval.
On a general interval
Suppose is a function on an interval
that may be infinite in one or both directions and may be open or closed at either end. Suppose
is a continuous function on all of
and that the derivative of
exists everywhere on the interior of
. Further, suppose
is an increasing function on
, i.e.:
Then, for all
in the interior of
. Further, there is no sub-interval of
such that
for all
in the sub-interval.
Related facts
Converse
Other similar facts
- Positive derivative implies increasing
- Zero derivative implies locally constant
- Negative derivative implies decreasing
Facts used
- Local maximum from the left implies left hand derivative is nonnegative if it exists
- Local minimum from the right implies right hand derivative is nonnegative if it exists
- Zero derivative implies locally constant
Proof
If is increasing on
, then every point in the interior of
is a point of local maximum from the left and local minimum from the right. Thus, by Facts (1) and (2), both the left hand derivative and the right hand derivative of
, if they exist, are nonnegative at any point in the interior of
. In particular, if the derivative itself exists at a point in the interior of
, then it must be nonnegative at that point.
It remains to show that the derivative is not zero on any sub-interval of . For this, note that by Fact (3), a derivative of zero forces the function to be constant on that sub-interval. This, however, contradicts the definition of an increasing function. We thus have the desired contradiction and we are done.