Good day, please help with this exercise, I am very confused, especially for the positivity and conjugate conditions.
Let $ V =C([0,1])$, the complex space of the continuous functions : $[0,1] \to \mathbb C$.
Show that the following is an inner product :
$\langle x, y\rangle = \int_0^1f(x)\overline{g(x)}dx$.
I guess you meant $\langle f,g\rangle=\int_0^1f(x)\overline{g(x)}dx$. You need to verify the axioms of inner product. The conjugate symmetry and linearity are really easy. The semi-positive-definiteness $\langle f,f\rangle\geq0$ is straightforward too. The only delicate point is to verify that $\langle f,f\rangle=0$ implies $f=0$:
Assume that $\langle f,f\rangle=0$ and that $f\neq0$ to get a contradiction. As $f\neq0$ there is some $x_0$ such that $f(x_0)\neq0$. As $f$ is continuous there must exist a closed interval $I$ containing $x_0$ such that $f(x)\neq0$ for all $x\in I$. This implies that $K:=\min_{x\in I}|f(x)|>0$ (which makes sense by the continuity of $|f|$), which leads to the contradiction $$ \langle f,f\rangle=\int_0^1|f(x)|^2dx\geq\int_I|f(x)|^2dx\geq K^2|I|>0. $$