Show $\langle f,g \rangle := \frac 1 {2\pi} \int^{\pi}_{-\pi} f(x) \overline {g(x)} dx$ define an inner product on complex vectorspace. Show $$\langle f,g \rangle := \frac 1 {2\pi} \int^{\pi}_{-\pi} f(x) \overline {g(x)} dx$$
define an inner product on the complex vector space of continuous functions.
I must establish $\langle f,g \rangle = \overline {\langle g,f \rangle}$, however I get to the point $$\langle g,f \rangle := \frac 1 {2\pi} \int^{\pi}_{-\pi} g(x) \overline {f(x)} dx =\frac 1 {2\pi} \int^{\pi}_{-\pi} g(x) \overline {f(x)} dx = \frac 1 {2\pi} \int^{\pi}_{-\pi} \overline {f(x)} g(x)  dx=\int^{\pi}_{-\pi} \overline {f(x) \overline {g(x)}}  dx$$, and I want to take the conjugate "out" the integral, but how ?
Also, how can I verify that $\langle f,f\rangle = 0 \iff f = 0$ ?
 A: It's ok to "take the conjugate out of the integral". After all, if $f(x) = u(x)+iv(x)$, where $u$ and $v$ are real-valued then
\begin{align}
\int_a^b \overline{f(x)}\,dx &= \int_a^b \big( u(x)-iv(x) \big)\,dx \\
&= \int_a^b u(x)\,dx - i\int_a^b v(x)\,dx \\
&= \overline{\int_a^b f(x)\,dx}.
\end{align}
Secondly if $\langle f,f \rangle = 0$, then 
$$
\int_{-\pi}^\pi |f(x)|^2\,dx = 0.
$$
For a slick way to see that this implies $f=0$, define
$$
g(t) = \int_{-\pi}^t |f(x)|^2\,dx.
$$
Clearly, $g(-\pi) = g(\pi) = 0$. On the other hand, the integrand is positive, so $g$ is increasing. Hence $g(t) = 0$ for all $t$, and conequently $g'(t) = 0$ for all $t$. On the other hand, the fundamental theorem of calculus implies that $g'(t) = |f(t)|^2$, so $|f(t)| = 0$ for all $t$, which in turn implies that $f(t) = 0$ for all t.
A: For any function $f$:
$$\int \overline{f(x)} dx = \overline{\int f(x) dx}$$
To see this, write $f(x) = g(x) + ih(x)$ with $g$ and $g$ being real-valued functions.
$$\int g(x) - ih(x) dx = \int g(x) dx - i \int h(x) dx$$
Now, to prove that $\langle f,f \rangle = 0 \implies f = 0$, notice that $\langle f,f \rangle$ is:
$$\frac{1}{2\pi} \int_{-\pi}^\pi f(x) \overline{f(x)} dx = \frac{1}{2\pi} \int_{-\pi}^\pi g(x)^2 + h(x)^2 dx$$
$g(x)^2 + h(x)^2$ is always positive, but the integral evaluates to $0$.  The rule for continuous functions is that if a nonnegative continuous function has integral $0$, then it is $0$ everywhere.  So $g(x)^2 + h(x)^2 = 0$.  I think you can fill in the rest of the details.
A: For the first question, note that $x$ is a real variable, so your complex integral is just defined as
$$
\int_{-\pi}^\pi f(x)\,\mathrm dx = \int_{-\pi}^\pi \operatorname{Re}(f(x))\,\mathrm dx + i\int_{-\pi}^\pi \operatorname{Im}(f(x))\,\mathrm dx.
$$
Thus conjugating $f$ under the integral is the same as conjugating the whole integral.
