Bijective & continuous -> differentiable?

[lolgarithms]

Is a bijective continuous function:[a,b]->[f(a),f(b)] differentiable?
I think it has to be.
continuity between two distinct values of f(a) and f(b): it got to take all the values between f(a) and f(b) at x in [a,b], by the intermediate value theorem.
if f is bijective, at [a,b], f(x) can't go up and then down at [a,b]. it has to be monotonically increasing for it to be bijective.
so a function can only be non-differentiable at a set of points of measure zero. (like the vertical tangent of f(x)=x^1/3 at 0)

 

[CompuChip]

The function f defined by:
f(x) = x when 0 <= x <= 1
f(x) = 2x - 1 when 1 <= x <= 2
on [0, 2] is a counter-example to the original statement "bijective & continuous => differentiable".

I suspect you can use this even to make a continuous bijection that is differentiable only on a set of measure zero, by taking a monotonously increasing function like above, of which the slope increases at every non-rational number.

 

[jostpuur]

so a function can only be non-differentiable at a set of points of measure zero.

Perhaps one could prove that using these:

http://en.wikipedia.org/wiki/Lebesgue's_decomposition_theorem

http://en.wikipedia.org/wiki/Absolutely_continuous

http://en.wikipedia.org/wiki/Singular_measure

I don't know those concepts very well, so I don't try to answer the question myself yet.

 

[HallsofIvy]

Is a bijective continuous function:[a,b]->[f(a),f(b)] differentiable?
I think it has to be.
continuity between two distinct values of f(a) and f(b): it got to take all the values between f(a) and f(b) at x in [a,b], by the intermediate value theorem.
if f is bijective, at [a,b], f(x) can't go up and then down at [a,b]. it has to be monotonically increasing for it to be bijective.
so a function can only be non-differentiable at a set of points of measure zero. (like the vertical tangent of f(x)=x^1/3 at 0)

Consider y(x)= x if 0\le x&lt; 1, 2x if 1\le x\le 2. That is a bijective from [0, 2] to [0, 4] and is continuous. It is NOT differentiable at x=1.

 

[CompuChip]

Consider y(x)= x if 0\le x&lt; 1, 2x if 1\le x\le 2. That is a bijective from [0, 2] to [0, 4] and is continuous. It is NOT differentiable at x=1.

It's not continuous, it has a jump at x = 1. If you shift it down by 1, like I did in my example, it works out.

 

[lolgarithms]

i mean... differentiable at all but a "small" (smaller than a real interval) set of points
like a finite set - like the one compuchip mentioned

 

[DrGreg]

First, I can think of a continuous bijection that fails to be differentiable on an uncountable set of measure zero, viz. the Cantor set. It is

f(x) = x + c(x)​

where c(x) is the Cantor function.

Second, that Wikipedia article links to Minkowski's question mark function, which is claimed to have a zero derivative on the rationals but is not differentiable on the irrationals.

 

[lolgarithms]

the question mark function -
ouch, i was wrong!