Find a continuous, strictly monotonic function f:R->R (R the set of real numbers) which is non-differentiable on a very dense set.

For this problem, we'll call a set of real numbers very dense if it intersects every interval [a,b] in an infinite, uncountable number of elements.

The following construction is the best I can do: Start with the linear function
l(x)=4x.  Then successively add functions which have "peaks" at regular
intervals.  The first function added is f_0(x).  It is piecewise linear between
the points ..., (-eps,0), (0,eps), (eps,0), (1-eps,0), (1,eps), (1+eps,0), ...,
where 0<eps<0.25.  f_1(x)=1/3*f_0(x-0.5), f_2(x)=1/9*f_0(2*x-0.5),
... f_n(x)=3^-n*f_0(2^(n-1)*x-0.5).  So the peaks of each successively added
function lie between those which are already there (perhaps this is
unnecessarily complicated, but it's much more beautiful that way ;).  The sum
of l(x) + f_0(x) + ... + f_n(x) converges to a continuous function as it
converges uniformly for n->infinity.  The maximally negative slope of f_n for
n>0 is -2^(n-1)/3^n, so the minimal slope (=lower bound for the difference
quotient) of the limit function is 4-1-1/3*sum(2/3)^n=2 > 0, so it is
monotonous.

What I'm not at all sure about is whether the limit function actually has a
noncountable number of non-differentiability points.  Each of the f_n has a
countably infinite number of such points.  But there are only countably many
f_n, and a countable union of countable sets is countable (like for instance
the set of points in the plane with whole coordinates).  True, the function
which I discussed with Steve Herman (see previous posts) is constructed
similarly, and is non-differentiable everywhere, on a noncountable set.  But as
far as I know, the proof of that relies on the fact that its "derivative" is
infinite everywhere, which is not the case for the function above.  Also, it
might seem that the limit function has more than countably many
non-differentiability points because the number of such points within a given
interval for the partial sums (up to f_n) increases exponentially, and because it
is known that 2^(countably infinite)=uncountably infinite.  But a countable
union of countable sets remains just countable, and the diagonal sequence
argument (proof of the non-countability of the reals) could not be applied to
the non-differentiability points of the limit function.

So the only way the given function can be the solution is for it to have two
different but finite limits of the difference quotient at certain (but not all
- see JLo's hint) irrationals.  Maybe I'm being stupid, but I don't see a way
to construct those limits, because the smaller peaks have to become ever more
flat to allow monotonicity.  If one wanted to construct the solution function
by successively adding non-differentiability points (without relying on
non-diff'ability at other points in the limit), one would have to add more than
countably many at at least one stage in the construction - probably at every
stage.

Posted by vswitchs on 2006-10-29 11:13:07