Jtdylktuy

For any positive real numbers $ alpha $ and $beta$, define

$f(x) = begin{cases} x^{alpha} sinfrac{1}{x^beta} && text{if $x in (0,1]$,}\
0 && text{if $x = 0$}end {cases}$

a) For a given $beta > 0$ , find all the values of $alpha$ such that $f'(0)$ exists.

b) For given $beta >0,$ find all the values of $ alpha$ such that $f$ is of bounded variation on $[0,1]$

My ANSWER: For $ a) $ $alpha ge beta$ then $f$ is bounded variation . now $ f'(0) $will exists if $alpha ge beta ge 0$

For $b)$ same condition as for $(a)$

Edit answer :$f$ has derivative $$displaystyle f^prime(x) = begin{cases} alpha x^{alpha-1} sinleft(dfrac{1}{x^beta}right) - dfrac{x^alpha}{x^{beta +1}} cosleft(dfrac{1}{x^beta}right) &text{on }(0,1], \\ 0 & text{if }x = 0.
end{cases}$$ Hence $$vert f^prime(x) vert le alpha x^{alpha-1} + x^{alpha - beta -1}$$ The integrals $int_0^1 x^{alpha-1}dx$ and $int_0^1 x^{alpha - beta -1}dx$ both converge for $1 < alpha < 1 +beta $. Hence $$V_0^1(f) le int_0^1 vert f^prime(x) vert dx$$ and $f$ is of bounded variation on $[0,1]$ as the RHS of the inequality is finite.

Is my answer correct ??? or incorrect ?? Please rectify it.

For the first part $f'(0)$ exists if $displaystylelim_{hrightarrow 0^+}dfrac{f(0+h)-f(0)}{h}$ exists. Without loss of generality fix $beta>0$, we have the limit $$displaystylelim_{hrightarrow 0^+}dfrac{(0+h)^{alpha}sinBig(dfrac{1}{(0+h)^{beta}}Big)-0}{h}=lim_{hrightarrow 0^+}h^{alpha-1}sinBig(dfrac{1}{h^{beta}}Big) .$$ Using squeeze theorem we see the limit exists if $alphageq beta+1$.

$rule{17cm}{1pt}$

For the second part we calcualte $f'(x)$.

$$ f'(x)=begin{align}begin{cases}alpha x^{alpha-1}sinBig(dfrac{1}{x^{beta}}Big)-dfrac{beta x^{alpha}}{x^{beta+1}}cosBig(dfrac{1}{x^{beta}}Big),
&xneq0\0, &x=0end{cases}end{align}$$ If total variation is finite we say $f$ is of bounded variation. Also if $f$ is differentiable and its derivative is Riemann-integrable, its total variation is given by

$${displaystyle V_{a}^{b}(f)=int _{a}^{b}|f'(x)|,mathrm {d} x.} .$$

$$V_0^1(f)=int_0^1|f'(x)|dxleqalphaint_0^1x^{alpha-1}dx+betaint_0^1x^{alpha-beta-1}dx=1+dfrac{alpha}{alpha-beta} $$
(The above integrals Riemann integrable only if $alpha-1>-1$ and $alpha-beta-1>-1$). Thus the total variation $V_0^1(f)$ is finite if $alpha>0$ and $alpha>beta$.