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This is from a proof in Feller's Vol 2 (p.537). I am struggling to prove the following

and so it is clear that at some point $x_0$ either $|\Delta(x_0+)|=\eta$ or $|\Delta(x_0)|=\eta$.

I know that $F$ is increasing so all its one-sided limits exist and $G$ is continuous so all its limits alo exist. These functions are bounded so these limits are also bounded. Since $\Delta(x)$ is a difference of these two functions, its one-sided limits also exist everywhere. How to use the fact that $\Delta(x)$ vanishes at $\pm\infty$ to show that such a point $x_0$ exists? Obviously, $\Delta(x)=\sup_x\Delta(x)\text{ or }\inf_x\Delta(x)$.

This is a slightly different explanation of the same point in Chung's "A course in probability theory" (p.237). It says

Since $F-G$ vanishes at $\pm\infty$, there exists a sequence of numbers $\{x_n\}$ converging to a finite limit $b$ such that $F_n(x_n)-G_n(x_n)$ converges to $\eta$ or $-\eta$. Hence either $F(b)-G(b)=\eta$ or $F(b-)-G(b)=-\eta$.

I feel like I should use Bolzano–Weierstrass theorem here but unable to proceed. In particular, $f$ vanishing at $\pm\infty$ would mean $$ \eta=\sup\{ \Delta(x) |x\in (-\infty,+\infty)\}=\sup\{ \Delta(x) |x\in [-K,K]\} $$ for some a finite $K$. This would allow one to get a monotone converging subsequence of $x$. However, I still need to prove that $|\Delta()|$ either attains $\eta$ or that one of its jump points is equal to $\eta$.

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    $\begingroup$ Strictly speaking, the answer to the question in the title is NO. Take $f(x) = 0$ when $x\not = 0$ and $f(0)=1$. Then the supremum of $f$ is never $f(x_0^+)$ nor $f(x_0^-)$, or perhaps we should specify the definition of one-sided limits. $\endgroup$ Apr 16, 2023 at 19:55
  • $\begingroup$ @Gribouillis Thanks. I will correct the title as I gain more understanding of Feller's sentence. Would you be able to answer the questions in the main text? $\endgroup$
    – chuck
    Apr 16, 2023 at 19:58
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    $\begingroup$ Assuming $\eta>0$, if $x_n$ is a sequence such that $\lim |\Delta(x_n)|=\eta$, then $x_n$ must be bounded, otherwise it would have a subsequence tending to $\pm\infty$ and this is a contradiction because the limit would be 0. That answers the question of how to use the behaviour at infinity in the proof. $\endgroup$ Apr 16, 2023 at 20:13
  • $\begingroup$ @Gribouillis How do we know what limit $\lim|\Delta(x_n)|=\eta$ is (like right or left)? $\endgroup$
    – chuck
    Apr 16, 2023 at 20:15
  • $\begingroup$ @Gribouillis and how do you get such a sequence $\{x_n\}$ in the first place? $\endgroup$
    – chuck
    Apr 16, 2023 at 20:49

1 Answer 1

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Let’s assume that $\Delta$ is some function which vanishes at infinity and which has well defined left and right limits. Let $\eta=\sup_x |\Delta(x)|$.

If $|\Delta(x)|=\eta$ somewhere, then since $\Delta$ is continuous where defined, we’re done.

Otherwise, as the supremum, there’s some sequence of distinct points $x_n$ so that $|\Delta(x_n)|$ approaches $\eta$. Since $\Delta$ vanishes at infinity, $x_n$ is bounded. Splitting infinitely many points between whether $\Delta(x_n)$ are positive or negative, there must either be infinitely many of positive points or infinitely many negative.

Let’s assume there are infinitely many are positive (the negative case is identical) and remove the negative ones from the sequence. The remainder all must have $\Delta(x_n)$ approach $\eta$. This set of points $x_n$ is infinite and bounded, so by Bolzano Weistrauss, it has at least one limit point. Call a limit point $x_0$. There’s some subsequence which converges to $x_0$.

As a limit point there’s some subsequence of $x_n$, call it $y_n$, such that $y_n$ converges to $x_0$. Either infinitely many are less than $x_0$ or infinitely many are greater. Let’s handle the first case (the other is identical).

Taking the subsequence of points less than $x_0$, we get some sequence $y_n$ which converges from the left to $x_0$ and for which the limit of $\Delta(y_n)$ is $\eta$. Thus, since $\Delta$ has well-defined one sided limits, $\Delta(x_0^-)=\eta$.

Going back to full generality, we have that either the left limit or right limit is equal to either $-\eta$ or $\eta$ at some point $x_0$.

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  • $\begingroup$ There is a third case if all the $x_n$ but a finite number of them are equal to $x_0$. Then we have neither $\Delta(x_0^+) = \eta$ nor $\Delta(x_0^-) = \eta$ but just $\Delta(x_0) = \eta$. $\endgroup$ Apr 17, 2023 at 7:01
  • $\begingroup$ @Eric I am trying to make the answer as detail as possible so bear with me. Would you answer (include it in your answer) the following i) the result and conditions required for "As the supremum, there’s some sequence $x_n$ so that $|\Delta(x_n)|$ approaches $\eta$." to hold, ii) which Pigeonhole theorem are you using and why is it needed? $\endgroup$
    – chuck
    Apr 17, 2023 at 8:25
  • $\begingroup$ Good point. Either $|\Delta(x_n)|$ is exactly $\eta$ somewhere (and we’re done) or the values in the sup limit necessarily are all different as they converge. Pigeonhole is: if I have an infinite number of points partitioned into a finite number of sets, some partition is infinite. $\endgroup$
    – Eric
    Apr 17, 2023 at 22:22

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