Domain of Orthogonality of Legendre Polynomials?

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In my Numerical Analysis Course, under the topic Theory of Orthogonal Polynomials We are following the book Numerical Analysis by Kendall E Atkinson.

The problem is the author keeps saying "The Legendre Polynomials are orthogonal in $(-1,1)$"

According to me a polynomial is defined everywhere and is Orthogonal With respect to a Inner-product which might have some integral and limit. It makes no sense for me to talk about orthogonality in a Domain.

The course is combined for Physics and Mathematics students and I am a Mathematics Major. Each Physics major believes the polynomial are orthogonal only in the domain. for example Legendre polynomials are orthogonal in $(-1,1)$ only and you cant use them outside the $(-1,1)$ to study them or to find the coefficients using recursive relation outside $(-1,1)$ using Gaussian Elimination.

I want to know if there is meaning of Polynomials being orthogonal in a domain. If Yes, what does it mean?

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The orthogonality of Legendre polynomials is typically defined with respect to the inner product

$$ \langle p,q\rangle = \int_{-1}^1 p(x)q(x)\,dx $$

Thus, they are said to be orthogonal on $(-1,1)$. Of course, through a linear change of variables you could define them to be orthogonal on any bounded interval $(a,b)$.

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What it means is that they are orthogonal with respect to the "uniform" probability distribution on $(-1,1)$, i.e. the measure $m$ defined by $$ m(A) = \begin{cases} 1 & \text{if } A = (-1,1), \\ 0 & \text{if } A \text{ is disjoint from } (-1,1), \\ \cdots & \text{(see below for other sets.)} \end{cases} $$ The measure assigned by $m$ to intervals that are subsets of $(-1,1)$ is proportional to the length of the interval (and from the above you can see what the constant of proportionality is).

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"The Legendre Polynomials are orthogonal in (-1,1)"

The Legendre Polynomials are orthogonal with respect to the inner product $\langle f,g\rangle=\int_\mathbb R w(x)f(x) g(x)\, dx$ for a certain weight function $w(x).$