One of my problem set questions asks that given $f \in L([0,1])$ and $g$ be a finite-valued monotone increasing function on $[0,1]$, and given any interval $[a,b] \subset [0,1]$ it holds that
$$ \bigg| \int_a^b f(x) \ dx \bigg|^2 \leq [g(b) - g(a)] (b-a) $$
I'm asked to show that $f^2 \in L([0,1])$.
I'm stuck as I see how the term on the right sort of forms a pseudo rectangle but I do not see any way to get the term $f^2$.
Thanks
Let $F(x)=\displaystyle\int_{0}^{x}f(t)dt$, then $\left(\dfrac{F(x)-F(y)}{x-y}\right)^{2}\leq\dfrac{g(x)-g(y)}{x-y}$, now use the facts that $g'$ exists a.e. and is integrable and $F'(x)=f(x)$ a.e.
Note that for $x>y$, $F(x)-F(y)=\displaystyle\int_{0}^{x}f(t)dt-\int_{0}^{y}f(t)dt=\int_{y}^{x}f(t)dt$, so $|F(x)-F(y)|^{2}\leq(g(x)-g(y))(x-y)$ and hence $\dfrac{1}{|x-y|^{2}}\left|F(x)-F(y)\right|^{2}\leq\dfrac{g(x)-g(y)}{x-y}$. Now assume that $F'(y)=f(y)$ and $g'(y)$ exist, then $\lim_{x\rightarrow y^{+}}\left(\dfrac{F(x)-F(y)}{x-y}\right)^{2}\leq\lim_{x\rightarrow y^{+}}\dfrac{g(x)-g(y)}{x-y}$ and it follows that $|f(y)|^{2}=|F'(y)|^{2}\leq g'(y)$.