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I stuck with this question, can you help me please.

Is it exist $ \mu$ - Borel regular measure in $[0,1]$ so that to all polynomial $p$ one has:

$\int_{[0,1]}p(t)d \mu(t)=p'(0)$?

Thanks a lot!

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I don't understand: we have $\int_{[0,1]}x^2d\mu(t)=0$ hence $\mu(0,1)=0$ and $\mu=a\delta_0+b\delta_1$. We have $b=0$ with $p=x$ and $a=0$ with $p=1$. –  Davide Giraudo Jun 22 '12 at 17:04
1  
@Davide Lilly didn't say it was a positive measure. –  Byron Schmuland Jun 22 '12 at 17:11
    
@ByronSchmuland Right, in fact it's quite obvious it cannot be a positive measure. –  Davide Giraudo Jun 22 '12 at 17:12
    
Hint: For Borel regular measures $C([0,1]) \ni p \mapsto \int_0^1 p\, d\mu$ is $\|\cdot\|_\infty$-continuous. –  martini Jun 22 '12 at 17:18

3 Answers 3

up vote 7 down vote accepted

For such a measure $\mu$ we have, for $n=0$ and $n\geq2$,
$$\int_{[0,1]}t^n d \mu(t)=0$$ and hence $\int_{[0,1]}p(t) d \mu(t)=0$ for all polynomials without linear term.

The sequence $p_n(t)=1-\sum_{j=1}^n \left|{1/2 \choose j}\right| (1-t^2)^j$ converges to $t$ uniformly on $[0,1]$, which implies $$0=\int _{[0,1]}p_n(t)d \mu(t)\to\int _{[0,1]} t\,d \mu(t)=1; $$ a contradiction.

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By Riesz theorem there exist isometric isomorphism $$ I:M([0,1])\to C([0,1])^* :\mu\mapsto\left(x\mapsto \int\limits_{[0,1]}x(t)d\mu(t)\right) $$ between Borel $\sigma$-additive measures and bounded functionals on $C([0,1])$. You can check that linear functional defined on dense subspace consisting of polynomials $$ \hat{f}:P([0,1])\to \mathbb{C}: p\mapsto p'(0) $$ is not bounded. Contradiction.

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We can use Bernstein polynomials: if $f$ is continuous on $[0,1]$, its Bernstein polynomial of degree $n$ is defined as $$P_n(x):=\sum_{k=0}^n\binom nkx^k(1-x)^{n-k}f\left(\frac kn\right).$$ We can see that $$P'_n(0)=\frac{f\left(\frac 1n\right)-f(0)}n$$ and that $P_n$ converges uniformly to $f$ on $[0,1]$. Hence, if we assume $\mu$ finite, we should have that $$\int_{[0,1]}fd\mu=\lim_{n\to +\infty}\frac{f\left(\frac 1n\right)-f(0)}n.$$ But this limit doesn't need to exist, as the function $$ f(x)=\begin{cases}x\sin\left(\frac 1x\right)&\mbox{ if }x\neq 0,\\ 0&\mbox{ if }x=0. \end{cases}$$

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