Increasing and differentiable implies nonnegative derivative that is not identically zero on any interval

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Statement

On an open interval

Suppose f is a function on an open interval I that may be infinite in one or both directions (i..e, I is of the form \! (a,b), (a,\infty), (-\infty,b), or (-\infty,\infty)). Suppose the derivative of f exists everywhere on I. Suppose further that f is an increasing function on I, i.e.:

x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)

Then, \! f'(x) \ge 0 for all x \in I. Further, there is no sub-interval of I such that \! f'(x) = 0 for all x in the sub-interval.

On a general interval

Suppose f is a function on an interval I that may be infinite in one or both directions and may be open or closed at either end. Suppose f is a continuous function on all of I and that the derivative of f exists everywhere on the interior of I. Further, suppose f is an increasing function on I, i.e.:

x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)

Then, \! f'(x) \ge 0 for all x in the interior of I. Further, there is no sub-interval of I such that \! f'(x) = 0 for all x in the sub-interval.

Related facts

Converse

Other similar facts

Facts used

  1. Local maximum from the left implies left hand derivative is nonnegative if it exists
  2. Local minimum from the right implies right hand derivative is nonnegative if it exists
  3. Zero derivative implies locally constant

Proof

If \! f is increasing on I, then every point in the interior of I 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 f, if they exist, are nonnegative at any point in the interior of I. In particular, if the derivative itself exists at a point in the interior of I, then it must be nonnegative at that point.

It remains to show that the derivative is not zero on any sub-interval of I. 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.