Continuous function on R at certain values of a

[maximus101]

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

 

[Amok]

Did you try actually integrating f(x) ?

 

[mjpam]

Did you try actually integrating f(x) ?

Doesn't Riemann integration assume continuity?

 

[spamiam]

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]

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.

 

[maximus101]

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.