# Does there exist a bijective differentiable function $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$, whose derivative is not a continuous function?

Does there exist a bijective differentiable function $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$, whose derivative is not a continuous function?

$x^2\sin \dfrac{1}{x}$ is a good example for non-continuous derivative function, that will not work here, I guess.

• My hunch is that a differentiable bijection will have a continuous derivative. No proof yet. Sep 3 '17 at 11:26
• @HennoBrandsma what do you mean by "No proof yet."? Sep 3 '17 at 11:38
• I don't have a proof of it now, I might get an idea later. Sep 3 '17 at 11:47
• @HyobinLee Bijective?
– Did
Sep 3 '17 at 12:04

The function $$f(x):=x^2\left(2+\sin{1\over x}\right)+8x\quad(x\ne0), \qquad f(0):=0,$$ is differentiable and strictly increasing for $x\geq-1$, and its derivative is not continuous at $x=0$. Translate the graph of $f$ one unit $\to$ and eight units $\uparrow$, and you have your example.
• If translated right, it takes negative values, so it isn't a function $\mathbb{R}^{+} \to \mathbb{R}^{+}$. You want the translation $g(x) = f(x-2) + f(0)$. Sep 3 '17 at 19:20
• Mistake in my original comment. Should be $f(x-1) + |f(-1)|$. Note that $f(-1) \approx -6.8$, so shifting up by eight units is too much. Sep 3 '17 at 19:51
Given any function $f:\mathbb{R} \to \mathbb{R}$ with a bounded derivative (e.g., $|f'| \leq M$, $M>0$) discontinuous at some $x>0$ and with $f(0) = 0$, the function $g(x) = f(x) + 2Mx$ defined on $\mathbb{R}_{+}$ gives an example.
It's surjective since $g'(x) = f'(x) + 2M \geq M$ so $g$ is an upper bound of the linear map $h(x) = Mx$. It's injective since the derivative is positive.
We can consequently construct some very bad examples.For example, we can use this as our $f$ so that $f'$ is discontinuous on a set of positive measure.