Continuity of a function $f$ on a set $A$ appears less strong than $f$ being pointwise continuous on every $x$ on a set $A$.

Specifically, the two definitions differ for points at the boundary of $A$.

More precisely,

(1) A function $f$ with domain $D$ is continuous on a set $A \subseteq D$ if, for any $x$ in $A$ and $\epsilon > 0$, the exists a $\delta > 0$ such that for $\bf y \in A$, $|x - y| < \delta $ implies $|f(x) - f(y)| < \epsilon$, .

(2) A function $f$ with domain $D$ is continuous for every $x \in A$ if, for any $x \in A$ and $\epsilon > 0$, there exists a $\delta > 0$ such that for any $\bf y \in D$, $|x - y| < \delta $ implies $|f(x) - f(y)| < \epsilon$.

That is, for (2), the requirement that $y$ be in $A$ is dropped. So clearly, (2) is stronger than (1) in general.

For open sets, the two definitions are equivalent, in that we can always pick a $\delta$ small enough that such that $|x-y| < \delta$ implies $y$ is in $A$.

My question is the following: is it possible to strengthen the definition of Uniformly Continuity & local Lipschitz continuity on a set $A$ to be more in line with (2)?

That is, something like

(3?) A function $f$ with domain $D$ is pointwise continuous uniformly for every $x \in A$ if for any $\epsilon > 0$, the exists a $\delta > 0$ such that for any $y \in D$ and every $x \in A$, $|x - y| < \delta $ implies $|f(x) - f(y)| < \epsilon$.

(4?) A function $f$ with domain $D$ is locally Lipschitz continuous uniformly for every $x \in A$ if there exists a $K$ such that every $x \in A$, there exist a neighborhood $U_x$ of $x$ in $D$ so that $f$ is Lipschitz continuous on $U_x$ with constant $K_x < K$.

As you can see (3) and (4) are stronger than the usual definitions of uniform continuity on a set and local Lipschitz continuity.

Interestingly, if $A$ is compact, (4) is also stronger than Lipschitz continuity on $A$ since for every $x$ on the boundary of $A$, there also exists a neighborhood not contained in $A$ such that $|f(x) - f(y)| \le K|x - y|$ for $y$ in that neighborhood.

Do definitions similar to (3) and (4) appear in the literature? They would be quite useful for my work, in particular to allow the uniform convergence of $|f(x) - f(y)|$ when $y$ converges from outside $A$ to the boundary of $A$.

  • $\begingroup$ Where have you get definition (2) from? The proper definition of pointwise continuity is relative to a specific domain for the function. $\endgroup$ – Rob Arthan May 24 '18 at 21:11
  • $\begingroup$ In the first definition you have domain "D", then define f to be continuous on A, a subset of D. In the second you have domain "D" and define f to be continuous on D. You don't have to say that y is in D because the definition uses "f(y)". In order for that to exist, y must be in D. $\endgroup$ – user247327 May 25 '18 at 16:18
  • $\begingroup$ If $y \notin D$ you could have $|x - y| < \delta$ but then it is not true that $|f(x) - f(y)| < \epsilon$, because $f$ is not defined for $y$. $\endgroup$ – Guillaume F. May 25 '18 at 19:39

Initially, continuity of a function is defined at a single point. Once that definition is under our disposal, we can define "continuity of a function on a set $A$", by requiring that it would be continuous at every point of the set $A$. Therefore, the two notions are precisely equivalent: continuity on a set means exactly this: continuity at every point of the set.

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    $\begingroup$ Continuity depends on the domain of the function. If we define $f(x) = 0$ for $x < 0$ and $f(x) = 1$ for $x \ge 0$, then $f$ is continuous on $[0, 1]$ but not on $[-1, 1]$. $\endgroup$ – Rob Arthan May 24 '18 at 21:14
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    $\begingroup$ uniquesolution: I have seen people give this definition: ƒ is continuous over the closed interval [a,b] iff it's continuous on (a,b), the right-sided limit of ƒ at x=a is ƒ(a) and the left-sided limit of ƒ at x=b is ƒ(b). Clearly this definition is similar to (1) but not to (2). $\endgroup$ – Guillaume F. May 25 '18 at 21:15
  • $\begingroup$ @RobArthan You have presented a tautology, because the very first word of your sentence, namely "continuity", refers to a "continuity on a set", and not to continuity at a point. So yes, Continuity in a domain depends on the domain. But the function fails to be continuous in this or that domain ONLY because it fails to be continuous at this or that POINT. $\endgroup$ – uniquesolution May 28 '18 at 20:16
  • $\begingroup$ Your answer and your comment ignore the fact that the notion of "continuity at a point" depends on the domain. In the example in my comment $f$ is not continuous at $0$ when its domain is taken to be $[-1, 1]$, but it is continuous at $0$ when its domain is taken to be $[0, 1]$. $\endgroup$ – Rob Arthan May 28 '18 at 23:23

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