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I want to know if there is a name for this property: for given positive numbers $a,b$, we have an inequality $|f(x)-f(y)|<b$ whenever $|x-y|<a$.

Warning:

It looks like Lipshitz continuity, but this property does not imply the continuity; for example, EVERY bounded function with bound $M$ satisfies it when we set $a$ and $b$ such that $a>0$ and $b>2M$. Note that this property depends on the choice of $a$ and $b$. Maybe the existence of these constants is important.

Where did I see this?:

I just saw this in several exam problems. I thought it might be useful in numerical analysis, because making $b$ small means that this function can be considered to be a Lipschitz continuous function with small errors. So I guessed it should have a name.

If anyone knows any concept similar to this, please tell me. Thanks.

user680089
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  • It sounds like you are either talking about Lipschitz continuity $$ |f(x)-f(y)|\leq (b/a)|x-y|$$ so that $(b/a)$ is the Lipschitz parameter, or uniform continuity (where we fix $b$ and then choose $a$). – Michael Jul 03 '19 at 02:37
  • nope. this condition does guarantee any kinds of continuity. – user680089 Jul 03 '19 at 02:39
  • I don't understand: are you trying to come up with a name for this, or did you see this somewhere and forget its name? – scoopfaze Jul 03 '19 at 02:44
  • I have never seen the name of this. But I guessed that it would be useful somewhere, in spite of discontinuity, because it provides with a powerful estimate of the function. I just encountered it when I was solving a final exam problem. – user680089 Jul 03 '19 at 03:03
  • But this is not a useful definition of continuity. If I fix $b$ at 2, and $a$ at 1, then this says that the floor function is continuous. – Matt Jul 03 '19 at 03:16
  • I know. I thought if it is used, maybe it will be in numerical analysis. If b is sufficiently small, we might make a good approximation. – user680089 Jul 03 '19 at 03:24
  • @Matt I could definitely see uses for it. Let's say you designed an algorithm that takes a real number $x$ and produces another real number $f(x)$. Suppose $\varepsilon > 0$ is the machine epsilon for whatever machine you're coding it for, and there was some $a$ such that $|x - y| < a \implies |f(x) - f(y)| < \varepsilon$. If you wanted to $f(x_0)$ where $x_0$ is not necessarily rational, you just need to compute $f(x)$ where $x$ is a decimal approximation within $a$ of $x_0$, and $f(x)$ will be indistinguishable from $f(x_0)$. Seems potentially handy. – Theo Bendit Jul 03 '19 at 03:34
  • I saw this three or four times in exam-oriented problems, so I just thought it sould have a name. – user680089 Jul 03 '19 at 03:35
  • I'm pretty sure Michael is correct and you mean Lipschitz Continuity only, I've seen it in various numerical Analysis courses as well – Anvit Jul 03 '19 at 03:36
  • @TheoBendit Thank you. My first ally! – user680089 Jul 03 '19 at 03:37
  • @Anvit Nope. This definition definitely admits step functions as "continuous functions". – user680089 Jul 03 '19 at 03:40
  • @user680089 are you talking about your definition or Lipschitz definition? because step function is continuous in both – Anvit Jul 03 '19 at 03:43
  • @Anvit I know that Lipschitz continuity with constant 1 is used to verify the existence of fixed point in numerical analysis. But I'm asking totally different concept – user680089 Jul 03 '19 at 03:43
  • @Anvit of course my definition. Lipschitz continuous function is continuous, but mine is never continuous at all. Please see quotation marks on "continuity". – user680089 Jul 03 '19 at 03:44
  • I think my question was very confusingly written.. sorry for you – user680089 Jul 03 '19 at 03:50
  • @user680089 : Not only are there quotation marks around "continuity," there are also quotation marks around "fixed." To understand the question, one must interpret both sets of quotation marks. It sounded like you were trying to remember a certain type of continuity, for which Lipschitz or uniform seemed like closest matches. Apparently that is not what you meant. Now you are saying quotations on "continuous" means "not really continuous" and so one expects quotations on "fixed" to mean "not really fixed." – Michael Jul 03 '19 at 14:05
  • ok alright I'll edit. – user680089 Jul 03 '19 at 14:08
  • @Michael Thank you for pointing out the problem. But I am very disappointed to downvoters so let me tell you what I'd like. Remeber some people understood and upvoted the question before edit. I know I'm also at fault, but I hope you guys not to justify your misunderstanding. Actually it should have been clear if you gave a more attention. – user680089 Jul 03 '19 at 14:43
  • @user680089 : I never downvoted it, but now I have upvoted it. – Michael Jul 03 '19 at 15:01
  • @Michael I am ashamed... I got upset and lost rationality temporarily. I will be careful at asking questions. Thanks – user680089 Jul 03 '19 at 15:07

1 Answers1

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Not all properties are required to have names, we can just make up a name like $P(a,b)$. However, your property is similar to the concept of a "leaky bucket constrained" arrival process used in queueing theory.

Definition 1:

Given positive real numbers $a,b$, a function $f:\mathbb{R}\rightarrow\mathbb{R}$ has property P(a,b) if $|f(x)-f(y)|\leq b$ whenever $|x-y|\leq a$.

Definition 2:

A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is leaky bucket constrained with rate parameter $\lambda \geq 0$ and burst parameter $\sigma \geq 0$ (written $f(t) \sim leaky(\lambda, \sigma)$) if $$ |f(x)-f(y)|\leq \lambda |x-y| + \sigma \quad \forall x, y \in \mathbb{R}$$

From these two definitions we can prove the following:

  • $f(t) \sim leaky(\lambda, \sigma) \implies$ $f(t)$ has property $P(a,\lambda a+\sigma)$ for every $a\geq 0$.

  • If $f(t)$ has property $P(a,b)$ for some positive numbers $a,b$ then $f(t) \sim leaky(b/a, b)$.

Definition 2 is usually applied to nondecreasing functions that represent arrival functions, such as functions $N(t)$ that represent the number of jobs that arrive to a queue during a time interval $[0,t]$. Such $N(t)$ functions jump discontinuously when a new arrival occurs. If $N(t) \sim leaky(\lambda, \sigma)$ then any interval of size $T$ has at most $\lambda T + \sigma$ arrivals, and the long term arrival rate satisfies $$\limsup_{t\rightarrow\infty} \frac{N(t)}{t} \leq \lambda$$ Such processes are useful because, assuming each job has a fixed size, the worst-case size of the queue can be deterministically bounded (in terms of $\sigma$) whenever the service time of each job is a constant $1/\mu$ that satisfies $\mu\geq \lambda$.

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Michael
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  • Note: The deterministic queue bound that holds for $leaky(\lambda,\sigma)$ processes whenever $\lambda \leq \mu$ is very strong and cannot be obtained in queues with Poisson arrivals, such as M/M/1 or M/D/1. Indeed M/M/1 or M/D/1 queues cannot be deterministically bounded and they require $\lambda < \mu$ even to ensure the average queue size is bounded, which is more stringent than $\lambda \leq \mu$. – Michael Jul 03 '19 at 18:21