Trying to show that the system $\cos nx$ is complete in $L_2[0,\pi]$, I am following a proof that $(\sin nx)_n$ is a complete system, in a book by Eidelman who currently teaches this course, and there it states:

For $\overline{f}(x) =

\begin{cases}

f(x), & \text{if $0\le n\le \pi$} \\

-f(x), & \text{if $-\pi\le n< 0$}

\end{cases}$ it holds that $\int_{-\pi}^{\pi}{\overline{f}(x)dx}=0$, and because $\overline{f}(x)$

is odd, $\int_{-\pi}^{\pi}{\overline{f}(x)\cos(nx)dx}=0$. Later it's shown that $\int f(x)\sin nx=0$ in a similar manner, using the fact that an element that is orthogonal to a system is zero when the system is complete.

I am having trouble simply understanding why $\int_{-\pi}^{\pi}{\overline{f}(x)dx}=\int_{-\pi}^{0}{-{f}(x)dx}+\int_{0}^{\pi}{{f}(x)dx}=0$

- Isn't $\int_{-\pi}^{0}{-{f}(x)dx}=\int_{0}^{\pi}{{f}(x)dx}$?

I don't feel comfortable approaching him with little things, even if there might have been a typo in their book. I could really use a hand here.

The course is "Introduction to Hilbert Space and Operator Theory", just to make it clear what Mathematical tools are used.

Edit:

I overlooked that $\int_{-\pi}^{0}{-{f}(x)dx}=\int_{0}^{\pi}{{f}(-x)dx}$.

## Best Answer

You have quoted the definition of $\overline f (x)$ wrongly. Your definition does not make sense because $f(x)$ is not defined when $x <0$. The correct defintion should be $\overline{f}(x) = \begin{cases} f(x), & \text{if $0\le x\le \pi$} \\ -f(-x), & \text{if $-\pi\le x< 0$} \end{cases}$. With this definition I believe you can see why $\int_{-\pi}^{\pi} \overline f(x)dx=0$.