The integral here is defined in the Riemann-Stieltjes sense, the interval $[a,b]$ contains $\psi$'s support, and $\alpha<1$. $X$ is $\alpha$-Hölder continuous.

\begin{align*} \left|\int_a^b \psi(v)\mathrm{d}X(v)\right|&\leq \|\psi\|_{\infty}\lim\limits_{\left|\mathcal{P}\subseteq{[a,b]}\right|\to 0}\sum\limits_{i=0}^{n-1}\left|X(v_{i+1})-X(v_i)\right|\\ &\leq \|\psi\|_{\infty}\mu([a,b])^{1-\frac{1}{\alpha}}\lim\limits_{\left|\mathcal{P}\subseteq [a,b]\right|\to 0}\left(\sum\limits_{i=0}^{n-1}\left|X(v_{i+1})-X(v_i)\right|^{\frac{1}{\alpha}}\right)^{\alpha}, \end{align*}

where $\mu$ is the standard Lebesgue measure. Have I correctly used the result that, for $1\leq p<q\leq\infty$, where $A$ is a finite measure space and $f$ is measurable (with measure $\mu$),we have $$\|f\|_p\leq \mu(A)^{\frac{1}{p}-\frac{1}{q}}\|f\|_q$$ ?

  • 1
    $\begingroup$ Where did $n$ go in $\sum_{I=0}^{n}1^{...}$? $\endgroup$ Apr 12, 2017 at 14:03
  • 1
    $\begingroup$ I don't see how you're justifying what you have between the second and third line of your first long inequality. Don't omit steps there. $\endgroup$ Apr 12, 2017 at 21:57
  • 1
    $\begingroup$ I still don't get it. The variable of integration is $v.$ what is the variable $t?$ $\endgroup$
    – zhw.
    Apr 15, 2017 at 17:44
  • 1
    $\begingroup$ OK, but if $X$ is not increasing, then why is the sum in the second line positive? $\endgroup$
    – zhw.
    Apr 15, 2017 at 18:03
  • 1
    $\begingroup$ That can't be right. If $X(t)= |t|$ on $[-1,1],$ then the first sum is always $0.$ You may want to move the absolute values inside the sum. $\endgroup$
    – zhw.
    Apr 15, 2017 at 18:56

2 Answers 2


As far as I understand you are claiming that

$$\sum_{i=0}^n |a_i| \leq c\left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha} \tag{1}$$

for some absolute constant $c=c(\alpha)$ and $\alpha \in (0,1)$. If we choose $a_i := 1$ for all $i=0,\ldots,n$, this would imply

$$n = \sum_{i=0}^n |a_i| \leq c \left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha} = c n^{\alpha};$$


$$c \geq n^{1-\alpha} \xrightarrow[]{n \to \infty} \infty.$$

This shows that there cannot exist a constant $c=c(\alpha)$ such that $(1)$ holds for any sequence $(a_i)_{i=0,\ldots,n} \subseteq \mathbb{R}$, $n \in \mathbb{N}$.

This, in turn, means that your reasoning doesn't work since the third "$\leq$" doesn't hold true.

Remark: The reason why $(1)$ does not hold true is that $n$ can become arbitrary large. In fact, Jensen's inequality shows that there exists a constant $c=c(\alpha,n)$ such that

$$\sum_{i=0}^n |a_i| \leq c \left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha}$$

for any $(a_i)_{i=0,\ldots,n}$. As the example in the first part of my answer shows, the constant $c=c(n,\alpha)$ explodes if we let $n \to \infty$.

Example Consider $[a,b] := [0,1]$ and the deterministic process $X_t := t$. If we set $t_i := i/n$ for fixed $n \in \mathbb{N}$, then

$$X_b-X_a = 1 = \sum_{i=0}^n (X_{t_{i+1}}-X_{t_i})$$ and

$$\left( \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^{1/\alpha} \right)^{\alpha} = n^{\alpha-1} \xrightarrow[]{n \to \infty} 0$$

and therefore there cannot exist a constant $c>0$ such that

$$\left| \int_a^b dX_t \right| = |X_b-X_a| \leq c \left( \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^{1/\alpha} \right)^{\alpha}$$

for all $n \in \mathbb{N}$. So even in this very simple setting we cannot expect that we can bound $\left\|\int_a^b \psi(t) \, dX_t \right\|_{\infty}$ by the $1/\alpha$-variation of $(X_t)_t$.

  • $\begingroup$ in order for $a_i=X_{t_{i+1}}-X_{t_{i}}$ to be equal to some $k$ for each $i$, wouldn't the function $X_t$ have to be infinitely steep? Is there a way to use the fact that I'm only considering the function in some bounded interval $[a,b]$? $\endgroup$ Apr 15, 2017 at 11:03
  • 1
    $\begingroup$ @man_in_green_shirt You didn't mention any assumptions on boundedness in your question (it really helps very much if you state clearly at the beginning of your question what objects you are dealing with and what assumptions you have... here, it is not even clear whether $X_t$ is a stochastic process or just a deterministic function). However, boundedness does not help, see my edited answer. Do you need the estimate only pointwise (i.e. with some constant which possibly depends on $\omega \in \Omega$) or global? $\endgroup$
    – saz
    Apr 15, 2017 at 11:19
  • $\begingroup$ what I'm trying to do is to bound $\|\int_a^b\psi\mathrm{d}X_t\|$, where $\psi$ vanishes outside of $[a,b]$. I would like this bound to contain the $\frac{1}{\alpha}$-variation of $X_t$ $\endgroup$ Apr 15, 2017 at 11:24
  • 1
    $\begingroup$ @man_in_green_shirt I see. Do you assume that $\sup_{\omega} \sup_{t \in [a,b]} |X_t(\omega)|<\infty$? If not, what kind of assumption do you have on the boundedness of $X$? $\endgroup$
    – saz
    Apr 15, 2017 at 11:29
  • $\begingroup$ I realise I should've mentioned this straight away, but I'm assuming it's $\alpha$-Hölder continuous $\endgroup$ Apr 15, 2017 at 12:04

If i well understand your last comment, i will give an answer using basic propriety of Riemann-Stieltjes integral : \begin{eqnarray} \left|\int_{a}^{b} \varphi(t) dX_t(t) \right|&=& \left|\int_{a}^{b} X_t(t) d\varphi(t) \right| \rm{ because }\, \varphi \,\rm{ vanish\, in } \, a \, \rm{ and }\, b.\\ &\leq & \int_{a}^{b}\left| X_t(t)\right| |d\varphi(t)|\\ &\leq& c \left( \int_{a}^{b}\left| X_t(t)\right|^{1/\alpha} |d\varphi(t)| \right)^\alpha \end{eqnarray} Where $c=\left(\int_a^b |d\varphi(t)|\right)^{1-\alpha}$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .