Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
@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
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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $$ f(x)=\begin{cases}x\sin\left(\frac 1x\right)&\mbox{ if }x\neq 0,\\ 0&\mbox{ if }x=0. \end{cases}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.