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Does there exist a signed regular Borel measure such that

$$ \int_0^1 p(x) d\mu(x) = p'(0) $$

for all polynomials of at most degree $N$ for some fixed $N$. This seems similar to a Dirac measure at a point. If it were instead asking for the integral to yield $p(0)$, I would suggest letting $\mu = \delta_0$. That is, $\mu(E) = 1$ iff $0 \in E$. However, this is slightly different and I'm a bit unsure of it. It's been a while since I've done any real analysis, so I've forgotten quite a bit. I took a look back at my old textbook and didn't see anything too similar. If anyone could give me a pointer in the right direction, that would be great. I'm also kind of curious if changing the integration interval from [0,1] to all of $\mathbb{R}$ changes anything or if the validity of the statement is altered by allowing it to be for all polynomials, instead of just polynomials of at most some degree.


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One quick question: Aren't measures usually nonnegative? Surely if we restrict ourselves to nonnegative measures, the measure in question cannot exist as the integral over the indicator functions must vanish, right? – Miklos Aug 5 '12 at 15:47
up vote 6 down vote accepted

Yes there does. In fact you can take any $N+1$ distinct points in $[0,1]$ and have the measure supported there. Suppose your points are $x_1, \ldots, x_{N+1}$. Let $e_j(x)$ be the unique polynomial of degree $\le N$ with $e_j(x_i) = 1$ for $i=j$, $0$ otherwise: explicitly $$e_j(x) = \prod_{i\in \{1,\ldots,N+1\} \backslash \{j\}} \frac{x - x_i}{x_j - x_i}$$
Note that for any polynomial $p$ of degree $\le N$, $p = \sum\limits_{j=1}^{N+1} p(x_j) e_j$. So you can take $\mu = \sum\limits_{j=1}^{N+1} e'_j(0) \delta_{x_j}$ and get $$ \int_0^1 p \ d\mu = \sum_{j=1}^{N+1} e'_j(0) p(x_j) = p'(0)$$

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This works perfectly and is extremely clear, thank you! I wish I had thought of using those polynomials, which I've seen before in classes a while ago. I suspect that my question is false if we change it from "all polynomials less than degree $N$ " to simply all polynomials. – Tom Jan 8 '12 at 5:58
That's right, because there are polynomials $p$ with $p'(0)$ arbitrarily large and $|p(x)| \le 1$ for all $x \in [0,1]$. – Robert Israel Jan 8 '12 at 7:03

As long as you're only interested in polynomials of degree less than or equal to some fixed $N$, then you can take $\mu$ itself to be a polynomial times $dx$ because you're looking at a finite dimensional inner product space. In this case, you can also regard it as a measure on the line which is supported on an interval.

My guess is that if you want to get polynomials of arbitrary degree, it won't be possible. A proof of that may proceed by something like RobJohn's suggestion -- if the measure were compactly supported, then, by Stone-Weierstrass, its values on continuous functions (and hence Borel sets) would be determined by its values on polynomials. Now the measure you're looking for is assumed to have finite moments of any order, but not necessarily compactly supported.

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I think you're right. I have little faith that it will work if we change it to polynomials of arbitrary degree. Thanks for the suggestions! – Tom Jan 8 '12 at 6:02

If you want a quick, lazy answer without any computation:

Consider the Banach space $C([0,1])$ and its $n+1$-dimensional subspace $\mathcal{P}_n$ consisting of all polynomials of degree at most $n$. The map $\ell : \mathcal{P}_n \to \mathbb{R}$ defined by $\ell(p) = p'(0)$ is linear, so since $\mathcal{P}_n$ is finite dimensional, $\ell$ is continuous. By the Hahn-Banach theorem it has a continuous linear extension $\tilde{\ell} : C([0,1]) \to \mathbb{R}$. By the Riesz representation theorem there is a signed regular Borel measure $\mu$ on $[0,1]$ such that $\tilde{\ell}(f) = \int f\,d\mu$, and for polynomials $p$ we have $\int p\,d\mu = \tilde{\ell}(p) = \ell(p) = p'(0)$.

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Yes, there are any number of nice choices. If you take $d\mu(x)$ to be a polynomial (times $dx$), say $\sum_{i=0}^{n}b_i x^{i} dx$, and integrate it against $p(x)=\sum_{i=0}^{n} a_i x^i$, your requirement is that $$ \int_{0}^{1} p(x)d\mu(x) = \int_{0}^{1} \sum_{i,j=0}^{n} a_i b_j x^{i+j} = \sum_{i,j=0}^{n}\frac{a_i b_j}{i+j+1} $$ must be equal to $a_1$. As a matrix equation, this is $$\langle a|\mathcal{\hat{H}}|b \rangle = \langle a | 0,1,0,0,\dots\rangle = \langle a | e_2 \rangle,$$ where $\mathcal{\hat{H}}$ is the Hilbert matrix of order $n+1$. The solution is clearly $b=\mathcal{\hat{H}}^{-1}|e_2\rangle$, or $b_i = (\mathcal{\hat{H}}^{-1})_{i+1,2}$. This has a known closed-form solution, $$ b_{i} = (-1)^{i+1}(i+2)(i+1)^2{{n+i+1}\choose{n-1}}{{n+2}\choose{n-i}} $$ (barring any transcription and renumbering errors), which is interesting in that the desired coefficients are all integers.

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