# Functions orthogonal to x^n

I would like to ask if there are any set of functions $u_n(x)$ which is orthogonal to $x^n$? i.e.:

$\int_0^1 x^n u_m(x) dx = \delta_{n,m}$

Edit: For clarification, this question asked for all non-negative integer m and n.

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You should be able to prove easily that a continuous function which is orthogonal to all xn is zero. Using that, you can answer your question. (This applies to continuous functions; if you want other kind of regularity, well, you'll have to adapt the argument) –  Mariano Suárez-Alvarez Jul 18 '10 at 15:17
the answers seem to reflect some confusion over whether the orthogonality is also over all n or is fixed for each n. –  Suresh Venkat Jul 19 '10 at 3:01

If $f \in L^2([0, 1])$ and $\int_0^1 x^n f(x)\, dx=0$ for all $n\ge N$ where $N$ is a nonnegative integer then $f$ is zero almost everywhere. To see this note that $x\mapsto x^N f(x)$ is an $L^2$ function orthogonal to all polynomials, and the polynomials are dense in $L^2([0,1])$. So the answer to your question is "no" for $L^2$ functions.

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Thank you for you answer. A follow up question: what if the interval changes? maybe to [-1,1] , or other interval? –  Ross Tang Jul 21 '10 at 4:24
The argument remains exactly the same for any bounded interval. –  Robin Chapman Jul 21 '10 at 5:20
Thank you very much. –  Ross Tang Jul 22 '10 at 15:53

The answer is no, and the main reason is that $\{x^n\}_{n = 0}^{\infty}$ form a total set in $L^2([0,1])$ so the set of their finite linear combinations is dense. But $\int x^n dx > 0$ for $n \geq 0$.

I believe, the best one can do is apply Gram--Schmidt to $x^n$ and obtain a sequence of polynomials $p_n$ (the orthogonal polynomials) of degree $n$ such that $$p_n \perp x^m,\quad m > n.$$ Here $f \perp g$ means $\int f(x) g(x) dx= 0$.

However, the notion of "best" here is not well-defined. It's just the usual choice.

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I'll answer this question for the interval [-1,1], but you should be able to extrapolate/specialize it to your case.

Let me define the inner product $\langle f,g\rangle=\int_{-1}^1 f(x)g(x) dx$ and normalized Legendre polynomials $P_m(x) = \sum_{n=0}^m p_{mn}x^n$ such that $\langle P_m,P_{m'}\rangle=\delta_{mm'}$. Note that $p_{mn}=0$ for $m>n$, so that the matrix $P$ with coefficients $p_{mn}$ is upper triangular. Let $Q$ be the inverse matrix, with components $q_{nm}$, that is, $x^n = \sum_{m=0}^n q_{nm}P_m(x)$. Clearly, $Q$ is also upper triangular. The triangularity matters because $Q$ can be found by explicit calculation, even though both $P$ and $Q$ are infinite dimensional.

It easily follows that $\langle x^n,P_m\rangle = q_{nm}$. It immediately follows that $\langle x^n,\sum_m p_{mn'} P_m\rangle = \sum_m q_{nm}p_{mn'} = \delta_{nn'}$. In other words, the functions you want are $u_{n}(x) = \sum_{m=0}^n p_{mn} P_m(x)$.

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These $u_m$ will satisfy $\int x^n u_m(x) dx=\delta_{m,n}$ for $n\le m$ but not necessarily for $n>m$. –  Robin Chapman Jul 19 '10 at 6:21

Derivatives of the Dirac distribution (appropriately normalized).

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I believe the question was about functions, not distributions. –  S. Carnahan Dec 7 '10 at 2:54