$ \int_{0}^{1}|f(x)|^{2}dx \leq \bigg|\text{Re}\int_{0}^{1}xf(x)\overline{f^{\prime}(x)}dx\bigg|. $

Question

Consider $f \in L^{2}(0,1)$ and $f^{\prime} \in L^{2}(0,1)$ with $f(1) =0$. Then

$$ \int_{0}^{1}|f(x)|^{2}dx \leq \bigg|\text{Re}\int_{0}^{1}xf(x)\overline{f^{\prime}(x)}dx\bigg|. $$

Note that $$ (x[f(x)]^{2})^{\prime} = f(x)^{2} + 2xf(x)f^{\prime}(x) $$ Soon, how $f(1) = 0$, then $$ \int_{0}^{1}|f(x)|^{2}dx \leq 2\bigg|\int_{0}^{1}xf(x)\overline{f^{\prime}(x)}dx\bigg|. $$ And now? And $\text{Re}$?

Answer 1

Since $|f|^2 = f\,\overline f$, just replace the first equation in your reasoning by $$ (x \,|f|^2)' = |f|^2 + x\, f' \,\overline f + x\,f\,\overline{f'} = |f|^2 + 2\,x\,\mathrm{Re}(f\,\overline{f'}). $$