# Continuous function on R at certain values of a

If we suppose $$a$$ > 0 is some constant and f: R $$\rightarrow$$ R is

f(x) = $$|x|^a$$ sin(1/x) if x $$\neq$$ 0

and

f(x) =0 if x=0

if we let F(x) := f '(x) for x $$\neq$$ 0 and F(0) :=0. For what values of
$$a$$ is F a continuous function in R

Did you try actually integrating f(x) ?

Did you try actually integrating f(x) ?

Doesn't Riemann integration assume continuity?

Doesn't Riemann integration assume continuity?

Definitely not. For instance, the integral of a step function is certainly defined. Lebesgue also proved that a function is Riemann integrable if and only if its set of discontinuities has measure zero (http://en.wikipedia.org/wiki/Riemann_integral), so for instance a countable number of discontinuities is acceptable.

HallsofIvy
Homework Helper
Why are you talking about integrating? The problem asked about the derivative of that function. F is clearly continuous for all x other than 0 so the only question is the derivative at x= 0.

For x> 0,
$$\frax{f(x)- f(0)}{x}= \frac{x^a sin(1/x)}{x}= x^{a-1}sin(1/x)$$
and, for any a> 1, the limit, as x goes to 0, is 0. If $a\le 1$ the limit does not exist.

For x< 0 we can replace x with the positive number y= -x and, since sine is an odd function we have
$$\frac{f(x)- f(0)}{x}= \frac{-y^a sin(1/y)}{-y}= y^{a- 1}sin(1/y)$$
and have the same result as before.

Why are you talking about integrating? The problem asked about the derivative of that function. F is clearly continuous for all x other than 0 so the only question is the derivative at x= 0.

For x> 0,
$$\frax{f(x)- f(0)}{x}= \frac{x^a sin(1/x)}{x}= x^{a-1}sin(1/x)$$
and, for any a> 1, the limit, as x goes to 0, is 0. If $a\le 1$ the limit does not exist.

For x< 0 we can replace x with the positive number y= -x and, since sine is an odd function we have
$$\frac{f(x)- f(0)}{x}= \frac{-y^a sin(1/y)}{-y}= y^{a- 1}sin(1/y)$$
and have the same result as before.

Hey thanks, could you explain why at x=0 f(x) is not differentiable for any value of a?
I think it is because the limit does not exist, for any a, but I don't know how to prove it.