Let $f$ be absolutely continuous on $[0,a]$ for all $a>0$ and $f(0)=0$ the show that $\int_{0}^{x}|f(t)f'(t)|\leq\frac{1}{2}(\int_{0}^{x}|f'(t)|)^2$.

Question

Let $f$ be absolutely continuous on $[0,a]$ for all $a>0$ and $f(0)=0$ the show that $\int_{0}^{x}|f(t)f'(t)|\leq\frac{1}{2}(\int_{0}^{x}|f'(t)|)^2$.

How I did it:

Clear for $x=0$. Now we take the derivative of both sides noting that indefinite integrals are absolutely continuous. Thus we get $|f(x)f'(x)|\leq|f'(x)|\int_{0}^{x}|f'(x)| \iff |f(x)|\leq\int_{0}^{x}|f'(x)|$ But the last inequality is true as $|f(x)|\leq TVf[0,x]$.

Is this correct? IS there a different way to justify $|f(x)|\leq\int_{0}^{x}|f'(x)|$ than they way I did?

Answer 1

Your solution is correct, but it overlooks a passage: the inequality $|f(x)|\le TVf[0,x]$ is not true in general, where we instead have $|f(x)|\le TVf[0,x]+|f(0)|$. However, since $|f(0)|=0$ by hypotesis, the particular inequality is valid and we are done.

Another derivation of $|f(x)|\le \int |f'(x)dx$ is by the second fundamental theorem of calculus, $$f(x)=f(0)+\int_0^xf'(t)dt\\ |f(x)|=\left|\int_0^x f(t)dt\right|\le\int_0^x|f(t)|dt\\$$