Looking for ${f_n}$ such that $\int_0^1 (x-t)^{m-1}f_n(t) dt = \delta_{n,m}$ Good day,
I am wondering whether it is possible to find a sequence of functions $f_n$ such that
$$\int_0^1 (1-t)^{m-1}f_n(t) dt = \delta_{n,m}$$
for every $0<n,m$.
Thank you.
 A: As suggested by John Huges one might start by transforming the formulas. One defines $f_n(t) \equiv g_{n-1}(t) $, having so 
$$
\int_0^1 (1-t)^{m-1}g_{n-1}(t) dt = \delta_{n,m} = \delta_{m-1,n-1}
$$
then setting $g_{n-1}(t) \equiv h_{n-1}(1-t)  $
$$
\int_0^1 (1-t)^{m-1}h_{n-1}(1-t) dt = \delta_{m-1,n-1}
$$
Doing now substitution $x=1-t$ and renaming $(m-1)\to m$ and $(n-1) \to n$ one gets
$$
\int_0^1 x^mh_n(x) dt = \delta_{m,n}
$$
Let be more general and study
$$
\int_a^b x^m h_n(x) dt = \delta_{m,n}
$$
Let $P_n$ be an orthonormal set of function in the $L_2$ metrics on $(a,b)$. Let both, $x^m$ and $h_n(x)$ express in its terms:
$$
x^m = \sum_i a_{m,i}P_i(x)
$$
$$
h_n(x)  = \sum_j b_{n,j}P_j(x)
$$
Computation leads to
$$
\delta_{m,n} = \int_a^b x^m h_n(x) dt \\
= \int_a^b \sum_i a_{m,i}P_i(x)  \sum_j b_{n,j} P_j(x) dt \\
= \sum_i \sum_j a_{m,i}b_{n,j}  \int_a^bP_i(x) P_j(x) dt \\
= \sum_i \sum_j a_{m,i}b_{n,j}  \delta_{i,j} \\
= \sum_i a_{m,i}b_{n,i}\\
$$
The last can be written in matrix notation
$$
1=ab^T
$$
which has clear interpretation, $a$ and $b$ are inverse (up to transposition). So if $a$ gives coefficients for expanding $x^m$ in terms of $P_j$ then $b$ gives the coefficients for expanding $P_n$ in terms of $x^j$. This is just power expansion (Taylor series) for orthonormal functions $P_n$ (often known). So, the function with the delta property should be written
$$
h_n(x)=\sum_k w_{k,n} P_k(x)
$$
where $w_{k,n}$ are such that
$$
P_k(x) = \sum_j w_{k,j} x^j
$$
Now the bad news: numerical computation suggest that the sum for $h_n$ diverges in all cases I studied, so I believe the answer to the initial question is "no", such function do not exist. But I miss a rigorous proof.
For more details here is self-promoting link:
http://vixra.org/abs/1804.0264
