Partial integration of $\int_0^L \overline{f(x)}(-f''(x))dx$

Question

I want to show that $\int_0^L \overline{f(x)} (-f''(x))dx$ is nonnegative with $f'(0)=0,f'(L)=0$. I tried with partial integration and got

\begin{align} \int_0^L \overline{f(x)} (-f''(x))dx &= [\overline{f(x)}(-f'(x)]_0^L - \int_0^L \overline{f'(x)}(-f''(x))dx \\ &= \int_0^L \overline{f'(x)}f''(x)dx \\ &= [\overline{f'(x)}f'(x)]_0^L - \int_0^L \overline{f''(x)}f''(x)dx \\ &= -\int_0^L \overline{f''(x)}f''(x)dx \end{align}

But I really dont see that the integral is nonnegative. Thank you in advance.

NOTE: there should be a bar over f(x) but it's quite small.

Answer 1

You have made errors in your integration by parts. It should read $$\int_0^L\overline{f(x)}(-f''(x))dx=\overline{f(x)}(-f'(x))\Big|_{x=0}^L-\int_0^L\overline{f'(x)}(-f'(x))dx=\int_0^L|f'(x)|^2dx,$$ which certainly is non-negative.