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Let $f$ be a function defined on a $J=[0,1]$, which is $k$-times differentiable on $J$; i.e. $f\in C^k(J)$.

By the Riesz representation theorem we know that for any complex Radon measure $\mu$ on $J$ , we have the isometric vector space isomorphism between the space of complex Radon measures on $J$ and the space of bounded linear functionals, given by $$\mu\mapsto \int f \ d\mu.$$

Given a bounded linear functional $I$ (which we know is of the form $\int f\ d\nu$) on $C^k(J)$, I want to show that we can find a unique $\mu, a_0,\dots,a_{k-1}$ where $\mu$ is a complex Borel measure and the $a_i$ are constants, with

$$I(f) = \int_{J} f\ d\nu = \int_J f^{(k)}\ d\mu + \sum_{i=0}^{k-1} a_i f^{(i)}(0)$$

This is also found as an exercise in Folland, 7.27, if that is helpful. Any guidance would be greatly appreciated.

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1 Answer 1

up vote 2 down vote accepted

We write Taylor's formula at $x_0=0$: $$f(x)=\sum_{i=0}^{k-1}\frac{f^{(i)}(0)}{i!}x^i+\int_0^x\frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)dt.$$ We have \begin{align*} \int_J\int_0^x(x-t)^{k-1}f^{(k)}(t)dt&=\int_0^1\int_{[t,1]}(x-t)^{k-1}f^{(k)}(t)\nu(dx)dt\\ &=\int_0^1f^{(k)}(t)\left(\int_{[t,1]}(x-t)^{k-1}f^{(k)}(t)\nu(dx)\right)dt, \end{align*} so we put $a_i:=\frac{f^{(i)}(0)}{i!}\int_J x^id\nu(x)$ and $\nu=g\cdot\lambda$, with $g(t):=\frac 1{(k-1)!}\int_{[t,1]}(x-t)^{k-1}f^{(k)}(t)\nu(dx)$ and $\lambda$ the Lebesgue measure.

Applying $I$ to $x^i,i=0,\ldots,k$ we can see that $a_i$ are uniquely determined, and then we can see that so is $\mu$.

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Thank you, this is crisp. –  Eric Gregor Mar 9 '12 at 0:40

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