6
$\begingroup$

$D[0,1]$ be the space of real functions $x$ on $[0,1]$ that are right continuous and have left hand limit.

On the other hand $C[0,1]$ is set of all continuous real functions on that interval.

Considering the underlying topology is Skorohod topology. I try to show that the set $C[0,1]$ is nowhere dense in $D[0,1]$. How to do that?

One idea is,

Suppose consider the ball in $D[0,1]$ i.e. $B(f,\epsilon) = \{g\in D[0,1] | d(f,g) < \epsilon\}$, where $d$ is Skorohod metric. Now it is enough to show that $C[0,1] \cap B(f,\epsilon) = \emptyset$. Then we can conclude that $C$ is nowhere dense.

But how to show this claim?

$\endgroup$
4
  • $\begingroup$ @AugustoSantos In $D$ uniform norm doesn't make sense that's why Skorohod metric required. I think. $\endgroup$
    – Cantor_Set
    Commented Apr 27, 2022 at 14:47
  • 1
    $\begingroup$ $C$ is a proper closed linear subset of $D$. Thus it has empty interior. $\endgroup$
    – Mittens
    Commented Apr 27, 2022 at 17:47
  • $\begingroup$ @OliverDíaz how to show that $C$ is proper closed linear subset of $D$? $\endgroup$
    – Cantor_Set
    Commented Apr 28, 2022 at 1:53
  • 1
    $\begingroup$ The topology $d_0$ restricted to $C$ coincided with the uniform topology in $C$. $\endgroup$
    – Mittens
    Commented Apr 28, 2022 at 2:45

2 Answers 2

6
+50
$\begingroup$

An explicit development.

Claim 1. Let $g\in D\left[0,1\right]$ be a càdlàg function with a jump of size $J>0$ at $t\in \left(0,1\right)$, i.e., $J=\left|g(t)-\lim\limits_{x\rightarrow t-} g(x)\right|$. Then, $$\|h-g\|\geq \frac{J}{2}\tag{$\star$}\label{star},$$ for any $h\in\mathcal{C}\left[0,1\right]$.

Proof. Remark that $|h(t)-g(t)|+|h(t)-\lim\limits_{x\rightarrow t-} g(x)|\geq J/2$. Consider $\delta>0$ small enough so that i) [h is continuous] $|h(x)-h(t)|<\varepsilon/3$ for all $x\in\left(-\delta,\delta\right)$; ii) [$g$ is right-continuous] $|g(x)-g(t)|<\varepsilon/3$ for all $x\in\left(0,\delta\right)$; and iii) [$g$ has left-limit] $|g(x)-\lim\limits_{x\rightarrow t-} g(x)|<\varepsilon/3$ for all $x\in\left(-\delta,0\right)$. Then, $|g(x)-h(x)|\geq J/2-\varepsilon$ for all $x\in \left(-\delta,\delta\right)$. In other words, $\|h-g\|\geq J/2-\varepsilon$ for any $\varepsilon>0$ and thus, $\|h-g\|\geq \frac{J}{2}$.

Claim 2. Let $g\in \mathcal{D}\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$ be a discontinuous càdlàg function -- i.e., it has a positive jump $J=\left|g(t)-\lim\limits_{x\rightarrow t-} g(x)\right|>0$ for some $t$. Then, $\mathcal{C}\left[0,1\right]\cap B\left(g,r\right)=\emptyset$ for any $r<J/2$, where $J$ is the size of one of the jumps of $g$.

Proof. Let $f\in \mathcal{C}\left[0,1\right]$. Observe that: $$d(f,g)=\inf_{\lambda\in \Lambda} \left\{ \|\lambda-I\|\vee \|f\circ\lambda - g\| \right\}\geq \inf_{\lambda\in\Lambda}\|f\circ\lambda - g\|.$$

Since $f\circ \lambda$ is continuous for any $\lambda\in\Lambda$ then, from Claim 1., we have that
$$\|f\circ\lambda - g\|\geq \frac{J}{2}\,\,\,\forall{\lambda\in\Lambda}.$$ Therefore, $d(f,g)\geq \frac{J}{2}$, for any $f\in \mathcal{C}\left[0,1\right]$. In other words, $\mathcal{C}\left[0,1\right]\cap B(g,r)=\emptyset$ for any $r<J/2$.

Claim 3. $\mathcal{D}\left[0,1\right]\setminus\mathcal{C}\left[0,1\right]$ is dense in $\mathcal{D}\left[0,1\right]$.

Proof. Given any $f\in\mathcal{C}\left[0,1\right]$, just consider the sequence $g_n(x):= f(x)+(\frac{1}{n})1_{\left\{x\geq t\right\}}$ and clearly $g_n\in\mathcal{D}\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$ for all $n$. We have $\|g_n-f\|\overset{n\rightarrow \infty}\longrightarrow 0$ (which also implies convergence w.r.t. the Skorohod metric).

Answer to the question. Since $\mathcal{D}\left[0,1\right]\setminus\mathcal{C}\left[0,1\right]$ is dense in $\mathcal{D}\left[0,1\right]$ (Claim 3.), the result follows: i) any (nonempty) open set $\mathcal{O}$ in $\mathcal{D}\left[0,1\right]$ contains a discontinuous $g$; ii) $B(g,J/3)$ does not contain any continuous function (Claim 2.), where $J>0$ is a jump of $g$. Therefore, $\mathcal{C}\left[0,1\right]$ is not dense in $\mathcal{O}$. This concludes the proof.

$\endgroup$
9
  • $\begingroup$ In the last line that is $J/2$ right? $\endgroup$
    – Cantor_Set
    Commented Apr 28, 2022 at 14:31
  • $\begingroup$ @Cantor_Set: it should be anything (strictly) smaller than $J/2$ (I have chosen $J/3$). In other words, any continuous $f$ lies outside the ball (induced by the Skorokhod metric) of radius $K$ around $g$ whenever $K<J/2$. $\endgroup$ Commented Apr 28, 2022 at 15:15
  • $\begingroup$ BTW typesetting suggestion: You can use \| to get $\|$ (rather than ||, which gives $||$). (I think I heard there's yet another code that gives the same result as \| but is semantically preferred for enclosing operators—supposedly, || is to be used for parallel lines and the like—but I don't remember offhand what it is.) $\endgroup$
    – Brian Tung
    Commented Apr 28, 2022 at 16:28
  • $\begingroup$ Ahh yes, it was \lVert and \rVert, so \lVert x \rVert gives $\lVert x \rVert$. $\endgroup$
    – Brian Tung
    Commented Apr 28, 2022 at 16:30
  • 1
    $\begingroup$ @WhoKnowsWho: $S$ is nowhere dense in $M$ whenever for any nonempty open subset $O$ of $M$, $S$ is not dense in $O$. The result above follows since any nonempty open set in $D\left[0,1\right]$ contains a discontinuous càdlàg function $g$. I will update as soon as possible to reflect that. In other words, $D\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$ is dense in $\mathcal{D}\left[0,1\right]$: thus, i) consider any open ball $\mathcal{O}$; ii) it contains a discontinuous càdlàg $g$; iii) $B(g,J/3)$ does not contain any continuous function, where $J$ is the jump of $g$. $\endgroup$ Commented Apr 28, 2022 at 18:07
4
$\begingroup$

Your idea is just opposite. To show that $C[0, 1]$ is nowhere dense in $D[0, 1]$, you need to show that $cl(C[0, 1])$ has empty interior. This means, you would want to start with an arbitrary ball centered at some $f\in cl(C[0, 1])$ and then show that that ball is not contained in $cl(C[0, 1]).$ This is the idea. How do we implement it?

I think the best way is to follow what Olvier Diaz mentioned in the comment. Start with the observation that the $cl(C[0, 1])=C[0, 1]$. This follows from the observation that Skorokhod metric restricted to $C[0, 1]$ is the same as $\|\cdot\|_{\infty}$.

With this observation, our problem reduces to showing that $C[0, 1]$ does not contain a ball (i.e., it has an empty interior). It is a general fact that in a topological vector space, a closed proper subspace is nowhere dense. If you haven't seen this already, there is an easy proof (I am writing it adapted to our case).

Let $f\in C[0, 1]$ and let $\epsilon>0$ be arbitrary. Consider the ball $$B(\epsilon, f):=\{g\in D[0, 1]: \|f-g\|<\epsilon\}.$$ It suffices to show that there exists $h\in D[0, 1]\setminus C[0, 1]$ such that $h\in B(\epsilon, f)$. This would show that this ball is not contained in $C[0, 1]$. As the epsilon was arbitrary, we conclude that $C[0, 1]$ does not contain any ball (and hence any open set).

To construct such a function $h$, define $h(x)=f(x)-sgn(x-1/2)\epsilon/2$ where $sgn(x)=1$ if $x>0$ and $sgn(x)=-1$ if $x<0$ and $sgn(0)=0$. Clearly $h\in D[0, 1]\setminus C[0, 1]$ and $\|h-f\|= \epsilon/2<\epsilon.$ This completes the proof.

$\endgroup$

You must log in to answer this question.

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