Conjugate symmetry to prove inner product

Question

We have to show that $$\langle p,q\rangle=\int_a^b \overline{p(t)}q(t)$$ is an inner Product.

I (think) I know what to do: I have to prove linearity, conjugate symmetry and positive definiteness. I have proven linearity and positive difiniteness already but I have difficulties with proving conjugate symmetry. I've come so far (which is not very far): $$ \overline{\langle q,p \rangle } = \overline{\int_a^b \overline{p(t)}q(t)}$$

I know that it continues with $\overline{\int_a^b \overline{p(t)}q(t)}=\int_a^b \overline{\overline{p(t)}q(t)}= \int_a^b p(t)\overline{q(t)}=\langle p,q \rangle$ but I don't fully understand why $$\overline{\int_a^b \overline{p(t)}q(t)}=\int_a^b \overline{\overline{p(t)}q(t)}$$ that is the case.

It would be great if someone could explain! :)

Answer 1

By definition, for a complex function $f(t) = u(t) + iv(t)$ ($u$ and $v$ real) we have $$\int_a^b f(t) := \int_a^b u(t) + i \int_a^b v(t). $$

Thus $$\overline{\int_a^b f(t)} = \overline{\int_a^b u(t) + i\int_a^b v(t)} = \int_a^b u(t) - i\int_a^b v(t) = \int_a^b \overline{f(t)}.$$

Answer 2

For a complex-valued function $f$, we may always decompose it into its real and imaginary parts $$ f = u + iv $$ where $u$ is the real part of $f$, $v$ is the imaginary part of $f$, and $u$ and $v$ are both real-valued functions. Then $$ \int f := \int u + i \int v \implies \overline{\int f} = \overline{\int u + i \int v} = \int u - i \int v = \int \overline{f}, $$ where $\overline{f} = u - i v$.