by nomadicmathematician
Last Updated October 09, 2019 17:20 PM

We denote by $H$ the subspace of $C([0,1])$ of functions $h$ such that $h(0)=0$, $h$ is absolutely continuous and its derivative $h'$ (which exists a.e.) satisfies $$\int_0^1 h'(s)^2 ds < \infty.$$

Then $H$ is a Hilbert space for the scalar product

$$(g,h) = \int_0^1 g'(s)h'(s)ds.$$

2) For any bounded measure $\mu$ on $[0,1]$, show that there exists an element $h$ in $H$ such that for every $f \in H$ $$\int_0^1 f(x)d\mu(x) = (f,h),$$ and that $h'(s) = \mu((s,1])$.

[Hint: The canonical injection of $H$ into $C([0,1])$ is continuous; use Riesz's theorem.]

Following the hint, we know that since $\int_0^1 f(x)d\mu(x)$ as a functional of $f$ is bounded linear, so it can be represented in the form $(f,h)$ for some $h \in H$. However I am stuck at showing that $h'(s)=\mu((s,1])$. $\mu((s,1]) = \int 1_{(s,1]} d\mu$, but this function does not belong to $H$. I would greatly appreciate any help.

We have \begin{align*} \int_{0}^{1}\left(\int_{0}^{s}\mu((t,1])dt\right)'(s)f'(s)ds&=\int_{0}^{1}\mu((s,1])f'(s)ds\\ &=\int_{0}^{1}\int_{0}^{1}\chi_{s<t\leq 1}d\mu(t)f'(s)ds\\ &=\int_{0}^{1}\int_{0}^{1}\chi_{0\leq s<t}f'(s)dsd\mu(t)\\ &=\int_{0}^{1}f(t)d\mu(t), \end{align*} so by the uniqess of Riesz, $\displaystyle\int_{0}^{s}\mu((t,1])dt=h(s)$ and hence $\mu((s,1])=h'(s)$.

We fix an $s<1$.

Let $(t_\epsilon)$ be the family of test functions (parametrized by $\epsilon >0$, possibly only with values $\epsilon =1/n$, $n>0$ natural,) which is piecewise linear, given by the values:

- zero on the interval $[0,s]$,
- one on the interval $[s+\epsilon, 1]$,
- linear on the remaining interval $[s,s+\epsilon]$, where it has the slope $1/\epsilon$, the jump of one, divided by the width of the interval. Then we have
*a.e.*in $s$ the equality:

$$ \begin{aligned} \mu(\ (s,1]\ ) &= \lim_{\epsilon\searrow 0} \mu(t_\epsilon) \\ &= \lim_{\epsilon\searrow 0} \int_0^1t_\epsilon'(u)\; h'(u)\; du \\ &= \lim_{\epsilon\searrow 0} \int_s^{s+\epsilon}\frac 1\epsilon\; h'(u)\; du \\ &= \lim_{\epsilon\searrow 0} \frac 1\epsilon (h(s+\epsilon)-h(s) \\ &=h'(s)\ . \end{aligned} $$

Updated March 07, 2017 03:20 AM

Updated August 24, 2018 05:20 AM

Updated August 02, 2017 14:20 PM

Updated March 28, 2018 22:20 PM

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