Function $f$ from $L^2[0,1]$ such that $\int \limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is polynomial.

Question

Is there function $f$ from $L^2[0,1]$ such that $\int \limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is:

a) Any polynomial of degree $\leq n$?

b) Any polynomial of any degree?

I know that the answer for b) is no: if $f$ is such function, then $xf(x)$ is orthoganal to any polynomial and thus orthogonal to any element of $L^2[0, 1]$. Therefore $f = [0]$.

I also understand that we can get a polynomial, which is orthoganal to $\{x, \dots, x^n\}$, using Gram–Schmidt process from $x$ to $x^{n+1}$. The problem is that I can't prove that such element is not orthoganal to $1$.

Thanks!

Answer 1

This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p \to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g \to \int fg$ for some $f \in L^{2}$.

Answer 2

In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $f\in\mathcal{L}^2\big([0,1]\big)$ for each given $n\in\mathbb{Z}_{\geq 0}$. This function $f$ is given by $$f(x)=f_n(x):=\sum_{k=0}^n\,(-1)^k\,(k+1)\,\binom{n+1}{k+1}\,\binom{n+k+1}{k+1}\,x^k$$ for all $x\in[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.