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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!

  • 0
    Hint: For Borel regular measures $C([0,1]) \ni p \mapsto \int_0^1 p\, d\mu$ is $\|\cdot\|_\infty$-continuous.2012-06-22

3 Answers 3

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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}$