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Let $D=D([0,1], \mathbb{R})$ be the space of cadlag functions $x$ with $x(0)=0$ and $x$ is continuous on $1$. If we endow $D$ with Skorokhod Metric, see:

http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g

for definition. Then I would like to know whether the following functions are uniformly continuous w.r.t. Skorokhod Metric:

$$\pi(x):=x(1/2)$$

$$S(x)=\sup_{0\le t\le 1}x(t)$$

$$L(x)=\int_0^1x(t)dt$$

I didn't find an answer in Convergence of Probability Measures, if someone knows the result please let me know. Thanks a lot!

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  • $\begingroup$ I know that $L$ is not uniformly continuous, and for the two other functions, I believe that they are not uniformly continous for some reason I can't see... $\endgroup$
    – CodeGolf
    Apr 9, 2014 at 8:14
  • $\begingroup$ If someone could give some examples, thx so much $\endgroup$
    – CodeGolf
    Apr 9, 2014 at 8:14
  • $\begingroup$ The first map $\pi$ is not even continuous: The indicator functions $x_n=I_{[0,1/2+1/n)}$ converge to $x=I_{[0,1/2)}$ (see Billingsley's book before example 12.1) but $\pi(x_n)=1$ does not converge to $\pi(x)=0$. $\endgroup$ Apr 9, 2014 at 8:54
  • $\begingroup$ Neither is the second continuous: $y_n=I_{[0,1/n)} \to 0$ but $S(y_n)=1$. $\endgroup$ Apr 9, 2014 at 9:22
  • $\begingroup$ $y_n$ can not converge to $0$ under Skorokhod Metric since for any $\lambda\in\Lambda$, we have $||y_n-0\circ\lambda||=1$ $\endgroup$
    – CodeGolf
    Apr 9, 2014 at 11:18

1 Answer 1

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Thx for Jochen Wengenroth, I think that $S$ is uniformly continuous: Let $\Lambda$ denote the collection of all strictly increasing functions from $[0,1]$ to $[0,1]$. Then for any $x, x'\in D$, we have for every $\lambda\in\Lambda$,

$$\sup_{t\in[0,1]}x(t)-\sup_{t\in [0,1]}x'(t)=\sup_{t\in[0,1]}x(\lambda(t))-\sup_{t\in [0,1]}x'(t)\le ||x\circ\lambda-x'||+||\lambda-I||$$

where $I(x)=x$ and $||\cdot||$ denotes the uniform norm. Hence

$$\sup_{t\in[0,1]}x(t)-\sup_{t\in [0,1]}x'(t)\le \inf_{\lambda\in\Lambda}\{||x\circ\lambda-x'||+||\lambda-I||\}$$

which gives

$$|\sup_{t\in[0,1]}x(t)-\sup_{t\in [0,1]}x'(t)|\le \inf_{\lambda\in\Lambda}\{||x\circ\lambda-x'||+||\lambda-I||\}=d(x,x')$$

Thus $S(x)$ is uniformly continuous.

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  • $\begingroup$ But I can not find a counterexample for $L$... $\endgroup$
    – CodeGolf
    Apr 9, 2014 at 9:05
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    $\begingroup$ Take a very rapidly oscillating function (say $1+\sin(nt)$). By slightly stretching the bits where it is close to $1$ and compressing the bits where it is close to $0$, you can change the integral by $O(1)$ while moving by only $O(1/n)$ in the Skorokhod metric. $\endgroup$ Sep 6, 2014 at 11:15

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