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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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    $\begingroup$ 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) $\endgroup$ Jul 18, 2010 at 15:17
  • $\begingroup$ the answers seem to reflect some confusion over whether the orthogonality is also over all n or is fixed for each n. $\endgroup$ Jul 19, 2010 at 3:01

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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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  • $\begingroup$ Thank you for you answer. A follow up question: what if the interval changes? maybe to [-1,1] , or other interval? $\endgroup$
    – Ross Tang
    Jul 21, 2010 at 4:24
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    $\begingroup$ The argument remains exactly the same for any bounded interval. $\endgroup$ Jul 21, 2010 at 5:20
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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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Derivatives of the Dirac distribution (appropriately normalized).

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  • $\begingroup$ I believe the question was about functions, not distributions. $\endgroup$
    – S. Carnahan
    Dec 7, 2010 at 2:54

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