Existence of functions with orthogonality property w.r.t polynomial

Question

Does there exist a set of functions $\{f_j\}$ and real numbers $a,b$ (potentially dependent on $i,j$) which have the property $\int_{a}^{b} x^i f_j(x) dx =A \delta_{ij}$ for some constant $A$? I don't have any idea as to how I could find such a set if it did exist and was wondering if anyone had any ideas.

Answer 1

Suppose we have linearly independent vectors $x_0,x_1,\ldots$ in a Hilbert space $H$. For each $0\leqslant n$, let $X_n=\text{span}\{x_k:k\leqslant n\}$. Let $P_n$ be the orthogonal projection onto $X_n$ and let $Q_n=I-P-n$ be the orthogonal projection onto $X_n^\perp$. For completeness, let $P_{-1}$ be the zero projection and $Q_{-1}$ be the identity operator.

Note that for $m\leqslant n$, $P_mP_n=P_nP_m=P_m$ and $Q_mQ_n=Q_nQ_n=Q_n$. Note also that, being orthogonal projections, these operators are self-adjoint.

Define $$w_n=P_{n-1}x_n$$ and $$v_n=Q_{n-1}x_n.$$ The sequence $(v_n)_{n=0}^\infty$ is orthogonal, so $\langle v_m,v_n\rangle=\|v_m\|^2\delta_{mn}$. Note that $$ \|v_m\|^2=\langle Q_{m-1}x_m,Q_{m-1}x_m\rangle = \langle x_m,Q_{m-1}^2 x_m\rangle=\langle x_m,Q_{m-1} x_m\rangle=\langle x_m,v_m\rangle.$$ Therefore, with $u_m=\frac{Q_{m-1}x_m}{\|v_m\|^2}$, $$\langle x_m,v_m\rangle = \frac{\langle x_m,v_m\rangle}{\|v_m\|^2}=1.$$ Furthermore, for $m<n$, $$\langle u_m,u_n\rangle = \frac{\langle Q_{m-1}x_m,Q_{n-1}x_n\rangle}{\|v_m\|^2\|v_n\|^2}=\frac{\langle Q_{n-1}x_m,x_n\rangle}{\|v_m\|^2\|v_n\|^2}=0.$$ Here we use the fact that $Q_{n-1}Q_{m-1}=Q_{n-1}$ and $Q_{n-1}\equiv 0$ on $X_{n-1}\ni x_m$.

Therefore the sequence $(u_m)_{m=0}^\infty$ has the properties we seek. How is this related to our examples? Fix any $a<b$. Let $H=L_2([a,b])$, the space of (equivalence classes Lebesgue a.e. classes of) of Lebesgue measurable functions on $[a,b]$ with finite second moment, with the inner product $\langle f,g\rangle=\int_a^b f(x)g(x)dx$ and norm $\|f\|=\sqrt{\langle f,f\rangle}$. This proves that for any $a<b$, functions $(u_m)_{m=0}^\infty$ exist satisfying $\int_a^b x^m u_n(x)dx=\delta_{mn}$.

How can we attempt to construct these functions explicitly? For each $n$, consider the $(n+1)\times (n+1)$ matrix $A_n=(\langle x_i,x_j\rangle)_{i,j=0}^n$. Here, $$\langle x_i,x_j\rangle = \int_a^b x^ix^j dx = \frac{b^{i+j+1}-a^{i+j+1}}{i+j+1}.$$ Note that for $i,j\leqslant m\leqslant n$, the $i,j$ term of $B_m$ is the same as the $i,j$ term of $B_n$, so the $b_{i,j}$ terms don't depend on $n$, except that $b_{i,j}$ isn't in $B_n$ until $i,j\leqslant n$.

Calculate $B_n=(b_{i,j})_{i,j=0}^n=A^{-1}_n$ and note that the orthogonal projection $P_{m-1}$ is given by $$P_{m-1}f = \sum_{i=0}^{m-1}\sum_{j=0}^{m-1} b_{i,j}\Bigl[\int_a^b f(x)x^i dx\Bigr]x^j.$$ In particular, we will have $$P_{m-1}x_m = \sum_{i,j=0}^{m-1}b_{ij}\Bigl[\int_a^b x^mx^i dx\Bigr]x^j.$$ From here, direct calculation can be made to get $Q_{m-1}x_m$ and, finally, $u_m$. In general, it will likely be difficult to calculate these things explicitly except in special cases.

A good starting point would be Legendre polynomials, which should already be the $v_m$ sequence above when $a=-1$. Depending on where you get the definition of the Legendre polynomials, they may or may not already be normalized (that is, they may or may not already be $u_m$). You Rodrigues' formula may give you the coefficients explicitly.