If we have a vector basis, when we perform Gram-Schmidt we then divide every vector by it's norm.
But if we have a polynomial basis, we can still write them as vectors (isomorphism) and perform Gram-Schmidt, but should we divide those vectors by their norm, since the norm of a polynomial does not make sense.
What would then be the correct way to orthonormalize polynomial basis.
Best Answer
Yes there is a norm in these spaces, given in general by an integral.
The most usual norm in the vector space of polynomials $\mathbb{R}_n[X]$ of degree less or equal to $n$ is derived from this dot product
$$(f|g):=\int_a^b f(t)g(t)dt$$
giving
$$(f|f)=\|f\|^2=\int_a^b f(t)^2dt \tag{1}$$
where $(a,b)$ may be $(0,1), \ (-1,1) $ etc...
A more general kind of dot product is:
$$(f|g):=\int_a^b f(t)g(t)w(t)dt \ \implies \ (f|f)=\|f\|^2=\int_a^b f(t)^2 w(t)dt$$
where $w$ is a weight function, i.e., a positive integrable function.
Let us take the case of
$$p_1(x)=(x-1), p_2(x)=(x^2-1) \in \mathbb{R}_2[x]$$
with $(a,b)=(-1,1)$ in (1).
They aren't orthogonal because
$$(p_1|p_2)=\frac12-1 = -\frac12 \ne 0 \tag{2}$$
Now let us keep $p_1$ fixed and look for $\lambda$ such that :
$$(p_1|(p_2+\lambda p_1))=0 \ \iff \ (p_1|p_2)+\lambda \|p_1\|^2=0$$ \iff $$\lambda=-\frac{(p_1|p_2)}{\|p_1\|^2}$$
As the numerator of $\lambda$ in (3) has been computed in (2), it remains to compute the denominator:
$$\|p_1\|^2=\int_{-1}^1 (x-1)^2 dx=\dfrac83$$
and so on if you have other vectors...
Another example here.