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.

  • 3
    $\begingroup$ Existence of such a "biorthogonal sequence" is equivalent to checking that $\{(1-t)^{m-1} \}$ forms a Riesz-Fisher sequence awgreen.people.clemson.edu/moment-riesz-equiv.pdf $\endgroup$
    – Walton
    Mar 5, 2018 at 15:22
  • $\begingroup$ It might be algebraically a little simpler to replace $(1-t)$ with $t$ in the integral; then when you find a set $f'_n$ with the required property, you can set $f_n(t) = f'_n(1-t)$. In this form, I recommend asking yourself what property $f_0$ has to have: you need $\int t^k f_0 = 0$ for $k > 0$, and $\int t^0 f_0 = 1$. So: $f$ has average value $1$, but integrates to $0$ against all higher-degree monomials. $\endgroup$ Mar 5, 2018 at 15:22
  • $\begingroup$ Is that a rhetorical question with an obvious answer? Well, I do not know.... I am just "hobbyist" so my overview might be quite small. I have rather tendency to thinking about some Gram–Schmidt process with polynomials maybe? Well, probably polynomials not... $\endgroup$
    – F. Jatpil
    Mar 5, 2018 at 15:30

1 Answer 1


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


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