|
Start with the Cantor function and denote it by C. This function is continuous, monotonic and it is differentiable everwhere, except on the Cantor set. In fact C is constant outside the Cantor set and therefore has a vanishing derivative. On points of the Cantor set, the left-side derivatives and the right-side derivatives do not exist and the difference quotients [C(x+h)-C(x)]/h and [C(x)-C(x-h)]/h approach +infinity when calculating the limit h->0.
Because the Cantor set is uncountable, the Cantor function already takes us 90% of the way. Unfortunately it is not a dense set (i.e. intersecting every interval) and thus certainly not very dense. Also, the Cantor function is defined on [0,1] only. To repair this, we need to extend it and to somehow "merge" it with a dense set, the most obvious example being the rational numbers. So let's take a counting r0, r1,.. of all rational numbers and define
Ci(x) := 0 for x < ri
Ci(x) := C(x - ri) for x in [ri, ri+1]
Ci(x) := 1 for x > ri + 1
In other words, the Ci's are constructed by shifting the original Cantor function by ri to the right along the x-axis and then extending it to the left and right as a constant function in order to have it well defined and continuous on ALL real numbers.
Now define
f(x) := Sum( 2-i * Ci(x) ), summing over all i = 0,1,..
I.e. we simply add up all shifted Cantor functions and use a squeezing factor 2-i to ensure we end up with something well-defined. Because our infinite sum converges uniformly and because all the summands are continuous, f is also continuous. The non-differentiability property of the Ci's has been extended for f to all rational shifts of the Cantor set, because the difference quotient [f(x+h)-f(x)]/h obviously approaches +infinity when x is inside one of the Cantor set shifts. This is because all the Ci's are monotonic, therefore by adding them all up the difference quotient will never become smaller by adding up extra Ci's. The union of all rational shifts of the Cantor set is obviously very dense, i.e. f has "enough" non-differentiability points. f is also monotonic because all the Ci's are. It is even strictly monotonic, because for every pair of values x1 < x2 there is a rational number ri that is strictly in between and we have Ci(x1) < Ci(x2), thus f(x1) < f(x2). But if you don't believe the "strict"-part, you could alternatively construct g(x):=f(x) + x to get something "even more monotonic".
Don't try to search for a function that is continuous, monotonic and nowhere differentiable: Every monotonic function is differentiable "almost everywhere", i.e. the set of points where a monotonic function is non-differentiable has vanishing lebesgue measure. See Wikipedia article on monotonic functions for this statement and see here for a proof.
|
