Abs Continuous Function w/ Unbounded Derivative on [a,b]

[glacier302]

What is an example of an absolutely continuous function on [a,b] whose derivative is unbounded?

I know that the function f: [-1,1] defined by f(x) = x^2sin(1/x^2) for x ≠ 0, f(0) = 0 is continuous and its derivative f'(x) = 2xsin(1/x^2)-2/xcos(1/x^2) for x ≠ 0, f'(0) = 0 is unbounded on [-1,1]. But this function isn't absolutely continuous...

Any help would be much appreciated : )

 

[Citan Uzuki]

We can construct another example along the same lines. Consider the following:

$$f(x)=\begin{cases} x^2 \sin (|x|^{-3/2}) & \text{if }x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$$

Computing the derivative yields:

$$f'(x) = \begin{cases} 2x \sin (|x|^{-3/2}) - \frac{3}{2} \operatorname{sgn}(x) |x|^{-1/2} \cos(|x|^{-3/2}) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$$

Since the derivative is integrable on [-1, 1], f is absolutely continuous on that interval.

 

[glacier302]

Thank you for your help. One question: How I do know that the derivative of that function is integrable? If f' were bounded, then the fact that it is only discontinuous at x = 0 would make f' Reimann integrable, and Reimann integrability implies Lebesgue integrability for bounded functions. But in this case, f' is not bounded, so how do I show that it is Lebesgue integrable?

 

[Citan Uzuki]

f' is obviously measurable, so you only have to show that the integral of the absolute value is finite. But we have:

$$\begin{align}\int_{-1}^{1} |f'(x)| \ dx & \leq \int_{-1}^{1} 2|x| |\sin(|x|^{-3/2})| + \frac{3}{2}|x|^{-1/2} |\cos(|x|^{-3/2})| \ dx \\ &\leq \int_{-1}^{1} 2 + \frac{3}{2} |x|^{-1/2} \ dx \\ & = 4 + \frac{3}{2} \int_{-1}^{1} |x|^{-1/2}\ dx \\ & = 4 + 3 \int_{0}^{1} x^{-1/2} dx \\ & = 4 + 6 \sqrt{x} \Big\vert_{0}^{1}\\ & = 10 < \infty \end{align}$$

 

[glacier302]

Aha. Thank you! I knew that if a bounded function is Reimann integrable then the Reimann integral and the Lebesgue integral of the function are equal. However, I always forget that this can be extended to unbounded functions as long as the Reimann integral is finite.

Thanks again!