I'm trying to prove that $||f||:=\sqrt{\langle f,f \rangle}$, where $\langle f_1,f_2 \rangle:=\int_{-\pi}^{\pi} f_1(t)\ \overline{f_2(t)}dt$, is a norm on the set of continuous functions, $C([-\pi,\pi],\mathbb{C})$.
Now, I proved that: $||f||\geq 0$, $||f||=0 \Leftrightarrow f=0$, $||\lambda f||=|\lambda|\cdot||f||$ but I'm stuck on the remaining one:
$||f_1+f_2||=\sqrt{\int_{-\pi}^{\pi}(f_1+f_2)(t)\ \overline{(f_1+f_2)}(t)\ dt}=\sqrt{\int_{-\pi}^{\pi}[|f_1(t)|^2+|f_2 (t)|^2+f_1(t)\overline{f_2(t)}+\overline{f_1(t)}f_2(t)]\ dt}$ . How can I show that $||f_1+f_2||\leq ||f_1||+||f_2||$?
The key is that the inner product satisfies Cauchy-Schwarz. You have \begin{align} \|f_1+f_2\|^2=\langle f_1+f_2,f_1+f_2\rangle&=\|f_1\|^2+\|f_2\|^2+2\text{Re}\,\langle f_1,f_2\rangle\\ \ \\ &\leq \|f_1\|^2+\|f_2\|^2+2\|f_1\|\,\|f_2\|=(\|f_1\|+\|f_2\|)^2. \end{align} The argument works for any sesquilinear inner product.