Jtdylktuy

For $a$, $b$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{a}sin(1/x^{b})$ for $0 < x leq 1$ and $f(x) = 0$ for $x=0$. Show that if $a> b$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $a leq b$, then $f$ is not of bounded variation on $[0, 1]$.

First, we see clearly that $f$ is continuous on $(0,1]$. Also, $f$ is continuous at $x=0$ because $lim_{x rightarrow 0}f(x)=0=f(0).$ Moreover, $f$ is differentiable over $(a,b)$ with $f'=ax^{a-1}sin(1/x^b)-bx^{a-(b+1)}cos(1/x^b)$ on $(0,1)$, and we see that $f'$ is continuous on $(0,1]$. I can not see the continuity for $f$ at x=0 becouse when $1>a>0$ we have a problem.

Any thoughts or ideas about that would be appreciated.

It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].