Concluding orthogonality

Question

In the space $P_1[0,2]$ with inner product defined as \begin{equation} \int_{0}^{2} p(x)q(x) \ dx \end{equation} Given \begin{equation} p(x) = 1-x \ \textrm{and} \ q(x) = 3+3x \end{equation}I know that they are NOT orthogonal in the given space. However, looking in $P_1$ with inner product defined as \begin{equation} a_0b_0 + a_1b_1 \end{equation} the two ARE orthogonal ( $1*3 + 1* -3 = 0)$. So my question is does this mean they would also be orthogonal in $P_2$ since the domain is open? $P_3$?

Answer 1

As you’ve already demonstrated in your question, orthogonality must be examined in the context of a specific inner product. You haven’t specified one for $P_2$ or $P_3$. You can’t simply use $a_0b_0+a_1b_1$ because it’s not positive-definite in those higher-dimensional spaces. On the other hand, if you extend it in such a way that it gives the same result when restricted to the subspace spanned by $1$ and $x$, then, yes, those polynomials will be orthogonal in those spaces, too. This is analogous to the situation in $\mathbb R^n$ with the standard Euclidean inner product. No matter how many zeros you tack on to $(1,0)$ and $(0,1)$ (relative to the standard basis, that is), the resulting vectors will be orthogonal.