Prove inner product of polynomials with complex coefficients

Question

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!

Answer 1

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 $V$.

Answer 2

You need to show:

1. $ \langle a p_1 + p_2 , q \rangle = a \langle p_1, q \rangle + \langle p_2, q \rangle $ for $ a \in \mathbb{C} $,
2. $ \langle p, q \rangle = \overline{\langle q, p \rangle} $,
3. $ \langle p, p \rangle > 0 $ for $ p \neq \mathbf{0} $.

Now, (1) is straight-forward from your definition. To prove (2), it is enough to show that $$ \int_0^1 p'(x) \overline{q'(x)} dx = \overline{\int_0^1 q'(x) \overline{p'(x)} dx} . $$ For a function $ f(x) $, denote the real part by $ f_r(x) $ and the imaginary part by $ f_i(x) $, so that $ f(x) = f_r(x) + i f_i(x) $ and $ \overline{f(x)} = f_r(x) - i f_i(x) $. Then,

$$ \int_0^1 p'(x) \overline{q'(x)} dx = \int_0^1 \left( p_r'(x) q_r'(x) + p_i'(x) q_i'(x) \right) dx + i \int_0^1 \left( - p_r'(x) q_i'(x) + p_i'(x) q_r'(x) \right) dx . $$

Similarly,

$$ \int_0^1 q'(x) \overline{p'(x)} dx = \int_0^1 \left( q_r'(x) p_r'(x) + q_i'(x) p_i'(x) \right) dx + i \int_0^1 \left( - q_r'(x) p_i'(x) + q_i'(x) p_r'(x) \right) dx , $$

and hence,

$\begin{align*} \overline{\int_0^1 q'(x) \overline{p'(x)} dx} &= \int_0^1 \left( q_r'(x) p_r'(x) + q_i'(x) p_i'(x) \right) dx + i \int_0^1 \left( q_r'(x) p_i'(x) - q_i'(x) p_r'(x) \right) dx \\ &= \int_0^1 p'(x) \overline{q'(x)} dx , \end{align*} $

which proves (2).

Next, to establish (3), consider any $ p \neq \mathbf{0} $. From the definition, $ \langle p, p \rangle = \| p(0) \|^2 + \int_0^1 \left\| p'(x) \right\|^2 dx $. If $ \langle p, p \rangle = 0 $, we must have $ \| p(0) \|^2 = 0 $ and $ \int_0^1 \left\| p'(x) \right\|^2 dx = 0 $, as both terms are non-negative. But $ \int_0^1 \left\| p'(x) \right\|^2 dx = 0 $ implies $ p(x) $ is constant over $ [0, 1] $, which coupled with $ \| p(0) \|^2 = 0 $ implies that $ p(x) = 0 $ for $ x \in [0, 1] $, i.e., $ p = \mathbf{0} $. Since we have considered $ p \neq \mathbf{0} $, we must have $ \langle p, p \rangle > 0 $, which proves (3).

Answer 3

Another approach is to reduce the problem to a known inner product.

First note that the evaluation map $E:\mathbb{P}^n \to \mathbb{C}^{n+1}$ defined by $Ep = (p(0),...,p(n))$ is linear and invertible.

Second note that $(p(0),...,p(n))^T = V (p_0,...,p_n)^T$, where the $p_k$ are the coefficients of $p(x) = \sum_{k=0}^n p_k x^k$ and $V$ is the appropriate Vandermonde matrix.

Hence the coefficient map $C:\mathbb{P}^n \to \mathbb{C}^{n+1}$ defined by $Cp =V^{-1} Ep$ is linear and invertible.

Now note that \begin{eqnarray} \int_0^1 p'(x)\overline{q'(x)} dx &=& \int_0^1 \sum_{j=1}^n j p_j x^{j-1} \overline{\sum_{k=1}^n k q_k x^{k-1}} dx \\ &=& \sum_{j=1}^n \sum_{k=1}^n jk p_j \overline{q_k} \int_0^1 x^{j+k-2} dx \\ &=& \sum_{j=1}^n \sum_{k=1}^n jk p_j \overline{q_k} H_{jk} \end{eqnarray} where $H$ is the Hilbert matrix. If we let $N=\operatorname{diag}(1,...,n)$ and $A=\operatorname{diag}(1, N^T HN)$ (the $1$ is for the $p(0)\overline{q(0)} = p_0 \overline{q_0}$ part) then we can write $\langle p, q \rangle = (Cp)^T A (\overline{Cq})$ where the latter is the (a) standard inner product on $\mathbb{C}^{n+1}$, where $A$ is positive definite. Linearity, conjugate symmetry and positive definiteness follow immediately from this and the fact that $C$ is linear and invertible.