Determine the formula and uniqueness of an inner product for a given orthonormal basis

Question

Let $P_n$ denotes the vector space of polynomials of degree at the most $n$. Let $\langle \enspace, \enspace \rangle$ be an inner product on $P_n$ with respect to which $\{1, x, x^2/2!, x^3/3!, \dotsc, x^n/n!\}$ is an orthonormal basis of $P_{n}$. Let $f(x) = \sum_i a_i x^i$ and $g(x) = \sum_i b_{i} x^i$ be two polynomials of $P_{n}$. Then which of the following option is correct?

a. $\langle f, g \rangle = \sum_i (i!)^2 a_i b_i$ defines one such inner product, but there is another such inner product.
b. $\langle f, g \rangle = \sum_i (i!)^2 a_i b_i$.
c. $\langle f, g \rangle = \sum_i (i!) a_i b_i$ defines one such inner product, but there is another such inner product.
d. $\langle f, g \rangle = \sum_i (i!) a_i b_i$.

My Working: I have found that options c and d are wrong as the vectors in the given orthonormal basis do not remain orthonormal in the inner product given in these two options. Option b is correct, as the given basis remains orthonormal in the inner product space given in this option. I am confused as I am not able to prove the uniqueness of this inner product space to accept or discard option a. Please help me to solve option a.

Answer 1

I will asuume that you are dealing with real polynomials. For complex polynomials the formula for $ \langle f, g \rangle$ would involve complex conjugates for $b_i$'s.

$f(x)= \Sigma_{i}$ $(i!a_{i})$$(x^i/{i!})$ and $g(x)= \Sigma_{i}$ $(i!b_{i})$$(x^i/i!)$ force $ \langle f, g \rangle$ to be $\Sigma_{i}$$(i!)^2$$a_{i}$$b_{i}$ by linearity of the inner product in each coordinate and by the definition of an orthonormal set.

Answer 2

An inner product is always uniquely determined by its values on all possible pairs of basis vectors. More specifically:

Let $p, q:V\to \Bbb R$ be two inner products on the finite dimensional real vector space $V$. Let $\{v_1, v_2, \ldots, v_n\}$ be a basis for $V$. Assume $$p(v_i, v_j) = q(v_i, v_j)$$for all $i, j$. Then $p=q$ (which is to say, $p(u, v) = q(u, v)$ for all $u, v\in V$).

(It is possible you can relax the constraints on $V$, but I didn't bother checking the involved subtleties today.)

The proof basically writes itself. Expand $u$ and $v$ in terms of the basis vectors, and use the bilinearity of the two inner products.

Using all products of all pairs of basis vectors (in order) is, in fact, the standard way to record the inner product into matrix form with respect to some basis, as we often do with vectors and linear transformations. If you have two vectors $u, v$ represented by column matrices (by abuse of notation also called $u$ and $v$), and an inner product represented by the matrix $A$, we have $$ \langle u, v\rangle = u^TAv $$ This matrix $A$ will always be symmetric and positive definite, and any symmetric positive definite matrix encodes a unique inner product in this way.