Implication of dot product $(Q_n(x), 1)\equiv \int_a^b {p(x)Q_n(x)dx} = 0$ in $L_2[a,b,p]$, where $Q_n(x)$ is a polynomial of $n$-th degree.

Question

Dot product $(Q_n(x), 1)\equiv \int_a^b {p(x)Q_n(x)dx} = 0$ in $L_2[a,b,p]$, where $Q_n(x)$ is a polynomial of $n$-th degree. According to my textbook this happens only if $Q_n(x)$ has at least one solution of odd order in $(a,b)$. I don't understand how it is derived. $p(x)$ is an arbitrary integrable on $[a,b]$ non-negative function so I don't see a connection. Any suggestions?

Answer 1

$Q_n(x)$ is a non-zero function, and so is $p(x)Q_n(x)$. If $\int_a^b p(x)Q_n(x)\,dx = 0$, then it cannot be the case that $p(x)Q_n(x) \geq 0$ for all $x$, nor that $p(x)Q_n(x) \leq 0$ for all $x$. Thus, there exist $x_1,x_2 \in (a,b)$ such that $p(x_1)Q_n(x_1) > 0$ and $p(x_2)Q_n(x_2) < 0$. Because $p \geq 0$, this means that $Q_n(x_1) > 0$ and $Q_n(x_2) < 0$.

By the intermediate value theorem, there must be a value $x_3$ within the open interval with boundaries $x_1$ and $x_2$ such that $Q_n(x_3) = 0$ at which the sign of $Q_n$ switches. Such a zero must have odd order.

Answer 2

What you have is a weighted inner product, I'm guessing you definition of $L^2[a,b,p]$ is $$L^2 ([a,b],p) := \left\{ f: [a,b] \rightarrow \mathbb{R}, \, f \text{ measurable}, \, \int_a^b p f^2 < \infty \right\} = L^2([a,b])$$ Here only because $[a,b]$ is compact. On $L^2([a,b])$ you have a canonical inner product $$\langle f,g \rangle := \int_a^b fg$$ But for each $p>0$ on $[a,b]$, you have an induced inner product with weight $p$ given by $$\langle f,g\rangle_p := \langle f,pg\rangle = \int_a^b pfg$$. You still have Cauchy-Schwarz since this is an inner product and everything that goes with it...

Since any polynomial belongs to $L^2([a,b])$ it is clear that the product is defined.

I suggest you check my answer on this question for more information.

For the solution of odd order, just show that there is a change of sign for $Q$ somewhere in $(a,b)$, it is equivalent to have a root with multiplicity $2m+1$