Borel Measure such that integrating a polynomial yields the derivative at a point

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.

Thanks!

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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

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.
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.