0

$g(x)=\sin(\frac{1}{x}) , g(0)=0$

Define $G(x)=\int _{0}^{x}g\left( t\right) dt$

Is $0$ a Lebesgue point of g? Prove or disprove it.

Frist , we know that $G'(0)=g(0)$ by A question about differentiation

so $\frac{1}{|B_r(0)|}\int_{B_r(0)}|g(z) - g(0)|\,dz =\frac{1}{|B_r(0)|}\int_{B_r(0)}|g(z) |\ $

I think that $\frac{1}{|B_r(0)|}\int_{B_r(0)}|g(z) |\ $ will not equal to $0$ , but I dont know how to prove it .

Community
  • 1
Mathboy
  • 91
  • Make a substitution $u = 1/z$. With $\int_c^{\infty} \frac{\lvert \sin u\rvert}{u^2},du$, it's easier to see how it behaves as $c\to +\infty$. – Daniel Fischer Jan 01 '17 at 19:21

1 Answers1

1

For $x>0,$

$$\int_0^x|\sin(1/t)|\,dt \ge \int_0^x \sin^2(1/t)\,dt$$ $$ = \int_0^x \frac{1-\cos (2/t)}{2}\,dt = \frac{x}{2} -\int_0^x \frac{\cos (2/t)}{2}\,dt. $$

Divide by $x$ to see

$$\tag 1 \frac{1}{x}\int_0^x|\sin(1/t)|\,dt \ge \frac{1}{2} - \frac{1}{x}\int_0^x \frac{\cos (2/t)}{2}\,dt.$$

The proof you know that $G'(0) = g(0)=0$ in the case of $g(t) =\sin (1/t)$ will work equally well with $g(t)=\cos (2/t).$ Thus the right side of $(1)$ $\to 1/2.$ Hence the left side of $(1)$ does not converge to $0,$ and therefore $0$ is not a Lebesgue point of $g.$

zhw.
  • 105,693