Let $f:[0,1] \to \mathbb R$ be Lebesgue integrable. Assume that $f$ is differentiable at $x=0$ and $f(0)=0$. Show that the function $g:[0,1] \to\mathbb R$ defined by $g(x)=x^{-3/2}f(x)$ for $x\in (0,1]$ and $g(0)=0$ is Lebesgue integrable.
I am studying for finals and am trying to wrap my head around this question. Would you use the Lebesgue-Vitali theorem that states, A bounded function $f:[a,b] \to\mathbb R$ is Riemann integrable if and only if it is continuous almost everywhere? or is this problem dealing with partitions, more specifically the mesh?? any help would be appreciated
Best Answer
You are just trying to show that it is integrable.
We know $g$ is measurable because the functions $x \mapsto x^{-{3 \over 2} } 1_{(0,1]}(x)$ and $f$ are.
Since $f$ is differentiable at $x=0$ and $f(0) = 0$, we know that for some $K$ and $\delta>0$ we have $|f(x)| = |f(x)-f(0)| \le K |x-0| = K|x|$ for any $x \in [0,\delta)$.
Hence we have $|g(x)| \le |x^{-{3 \over 2} } | K |x|= K {1\over \sqrt{|x|}}$ for all $x \in (0,\delta)$.
Furthermore, since $x \mapsto x^{-{3 \over 2} }$ is decreasing on $(0,1]$, we have $0\le |g(x)| \le ({\delta \over 2})^{-{3 \over 2}} |f(x)|$ for all $x \ge {\delta \over 2}$.
Combining, we see that \begin{eqnarray} \int |g| &\le& \int_{[0,{\delta \over 2})}|g| + \int_{[{\delta \over 2},1}|g| \\ &\le& \int_0^{\delta \over 2}K {1\over \sqrt{x}} dx + ({\delta \over 2})^{-{3 \over 2}} \int_{[{\delta \over 2},1} |f| \\ &=& \sqrt{\delta \over 2}K+{\delta \over 2}^{-{3 \over 2}} \int |f| \end{eqnarray} and so $g$ is integrable.