Consider set $\mathbb{P_2(\mathbb{C})}$. On this space we define the inner product:
$$\langle p, q \rangle = p(0)\overline{q(0)} + \int_0^1 p'(x)\overline{q'(x)} \, dx$$
where $p'(x)$ denotes the derivative of $p(x)$.
Now I have to show that $\langle p, q \rangle$ defines an inner product on $\mathbb{P_2(\mathbb{C})}$
Of course we verify all four properties of inner products. But I was a bit confused while verifying that $\overline{\langle p, q \rangle} = \langle q, p \rangle$ because of the complex conjugate.
Can I write $p(x)\overline{q(x)} = \overline{p(x)}q(x)$?
Also if the next step of proving that $\langle p, p\rangle > 0$, I have figured that we could write $p(x) = a_0+a_1x+a_2x^2+…+a_nx^n$ and have that
$$\langle p, p\rangle = (a_0+a_1x+a_2x^2+…+a_nx^n)(\overline{a_0}+\overline{a_1}x+\overline{a_2}x^2+…+\overline{a_n}x^n)$$
but I don't know how to move on from there.
Thank you for any kind of help!
Best Answer
I'll prove your form is inner product over $\mathbb C[x]=V$ (polynomial space with base field $\mathbb C$) by proving three properties: left linearity, Hermitian symmetry, positive definiteness.
First, I'll use notation such that $p(x)=\sum_0^np_ix^i\ (p\in V)$, simillar as $q,r\in V$. And consider $a\in\mathbb C$.
Part 1: Left Linearity $$ \langle p+q,r\rangle=(p+q)(0)\overline{r(0)}+\int_0^1(p+q)'(x)\overline{r'(x)}dx \\ =p(0)\overline{r(0)}+q(0)\overline{r(0)}+\int_0^1p'(x)\overline{r'(x)}dx+\int_0^1q'(x)\overline{r'(x)}dx \\ =\left(p(0)\overline{r(0)}+\int_0^1p'(x)\overline{r'(x)}dx\right)+\left(q(0)\overline{r(0)}+\int_0^1q'(x)\overline{r'(x)}dx\right) = \langle p,r \rangle+\langle q,r \rangle $$ $$ \langle ap,q \rangle=(ap)(0)\overline{q(0)}+\int_0^1(ap)'(x)\overline{q'(x)}dx=ap(0)\overline{q(0)}+a\int_0^1p'(x)\overline{q'(x)}dx \\ =a\left(p(0)\overline{q(0)}+\int_0^1p'(x)\overline{q'(x)}dx\right)=a\langle p,q \rangle $$
So, $\langle \cdot,\cdot \rangle$ satisfies left linearity.
Part 2: Hermitian Symmetry
Since $\overline{\alpha\overline\beta}=\overline\alpha\beta$... $$ \langle q,p \rangle=q(0)\overline{p(0)}+\int_0^1q'(x)\overline{p'(x)}dx=\overline{p(0)\overline{q(0)}}+\int_0^1\overline{p'(x)\overline{q'(x)}}dx \\ =\overline{p(0)\overline{q(0)}+\int_0^1p'(x)\overline{q'(x)}dx}=\overline{\langle p,q \rangle} $$
So, $\langle \cdot,\cdot \rangle$ satisfies Hermitian Symmetry.
Part 3: Positive definiteness
If $p\ne0$, then $p_0$ and $p'(x)$ cannot be both 0, because $p(x)=p_0+\int_0^xp'(t)dt$. So, $\|p_0\|$ and $\|p'(x)\|$ cannot be both 0. So...
$$ \langle p,p \rangle=p(0)\overline{p(0)}+\int_0^1p'(x)\overline{p'(x)}dx=\|p_0\|^2+\int_0^1\|p'(x)\|^2dx\ne0 $$
So, $\langle \cdot,\cdot \rangle$ is positive definite.
Your $\langle \cdot,\cdot \rangle$ is, so, inner product over $V$.