# standard symbol for a continuous function on some range

Sorry for what must be a very simple question, but Internet searches have failed me.

Is there a standard way to represent "continuous on some range"?

For instance, if I want to say

$g^{\prime}$ is continuous on $[a,b]$

, is there a way to represent the English words "is continouus on" by some standard symbol? I'm not looking for a definition of continuity, merely a standard symbol that represents the idea typographically.

[Edit, 2011.05.27:]

I've been asked to be more precise about the function. Actually, this example was taken from the first part of a published statement of the Substitution Rule for definite integrals, which I have been trying to put tersely on a flashcard using TeX. The full entry (James Stewart, Essential Calculus [N.p: Thomson, 2007] p.239 [Sec. 4.5]) reads:

If $g^{\prime}$ is continuous on $[a,b]$ and $f$ is continuous on the range of $u=g(x)$, then $$\int_a^bf\left(g(x)\right)g^{\prime}(x)~dx=\int_{g(a)}^{g(b)}f(u)~du$$

It's quite true that "range" appears in reference not to the interval $[a,b]$ but rather to $g(x)$. So I should further ask: is there a way to represent "continuous on the range" that is different from the representation "continuous on the interval", described by one of the commenters?

Really, as a philologist, I am boundlessly impressed at the typographical creativity of mathematicians and was sorry not to find a ready-made symbol anywhere for "continuous". Though I see that the notion of continuity is closely connected to notions of interval and boundedness.

-
The word range is usually where the function is "going to", while $[a,b]$ in your case is the domain is where the function is "coming from". –  Asaf Karagila May 27 '11 at 4:46
This is perhaps an unhelpful comment, but: your boxed statement fits neatly in a rather small box, so it would certainly fit on a standard-sized index card. Using too many abbreviations -- especially nonstandard ones -- can make things harder to read. I can't think of any clearer, more concise way to say "continuous on the range of $g$" than "continuous on the range of $g$": more symbols would certainly not help me here. –  Pete L. Clark May 27 '11 at 6:29
@texmad: by inserting the word "me" in my last comment, I hoped to convey that I was expressing a personal preference. The issue of exactly when the use of symbols makes mathematics easier to read and when it makes it harder is a complex one (and possibly of philological/linguistic interest, I would suggest). As someone who has been reading and writing latexed mathematics for a good while now, I have at least learned what I like and try to write in a way which pleases myself, at least. For me, what is written in the box above is essentially optimal. –  Pete L. Clark May 27 '11 at 6:46
As someone who is (like most mathematicians) also a teacher of mathematics, let me also say: all but the most precocious math students in the first half of their undergraduate career tend to use more symbols and fewer words (and especially, complete sentences) than I and most of my colleagues would like. A theorem is a sentence -- this is something that the logicians worked out formally, but it is equally true in the informal sense -- and taking out too many words runs the risk of interfering with the grammar of the sentence and thus its meaning. –  Pete L. Clark May 27 '11 at 6:50
(continued) One of the more distressing things to happen as an instructor is to ask (on an exam, say) for a definition or a theorem and get as an answer something which isn't even a well-formed sentence. I definitely call attention to grammatical and spelling mistakes on exams (and homework, when I grade it myself!) -- I don't take points off for it, but only because I think that would be badly received. In summary: mathematicians like words, perhaps more so than in many other technically-minded fields. (For instance, mathematicians use rather few acroynyms, especially compared to CS(!).) –  Pete L. Clark May 27 '11 at 6:59

I would propose $g'|_{[a,b]}\in C([a,b])$, because it sounds like you might want the possibility of the function $g'$ being defined on a larger set to remain open. In that case, we should be precise; we need to restrict $g'$ to a function whose domain is in fact $[a,b]$ before claiming it is an element of $$C([a,b])=\{f:[a,b]\rightarrow\mathbb{R}\mid f\text{ is continuous}\}.$$

-
It is striking to me that the one thing I had hoped to replace in the original expression, the English word "continuous", is still found in your equation! –  brannerchinese May 27 '11 at 5:48
@texmad: No, I proposed the expression $$g'|_{[a,b]}\in C([a,b])$$ Below that, I was merely pointing out the definition of $C([a,b])$ for the purpose of explaining why $g'|_{[a,b]}\in C([a,b])$ makes sense and $g'\in C([a,b])$ doesn't (necessarily) make sense. –  Zev Chonoles May 27 '11 at 6:01
I have to say that I think you are being overly cautious here. Notation is good if the reader has only one reasonable guess as to what it should mean. To me $f \in C([a,b])$ could mean only one thing. For instance, if $f: \mathbb{R} \rightarrow \mathbb{R}$ is the function which is $1$ at $x = 0$ and $0$ elsewhere, I would have no problem with ''$f \in C[1,10]$'' -- of course this means the restriction of $f$ to $[1,10]$: what else? –  Pete L. Clark May 27 '11 at 6:23
@Pete: I would probably say $f\in C[1,10]$ in real life, if nothing else to avoid an endless stream of $|\,$'s, but I would say it with a tiny bit of hesitation. I was already kind of "on alert" to the issue because the domain of $g'$ was not specified in the question; but furthermore, I felt that this kind of identification of $g'$ with its restriction to a subset was a "post-rigorous" conception of function, to use Terry Tao's terminology, –  Zev Chonoles May 27 '11 at 7:38
while the nature of the OP's question indicated to me that they are likely thinking in a "rigorous" mindset, so I wanted to give an answer suited to them. In fact I would have to admit that I haven't studied real analysis enough to confidently speak in a post-rigorous manner about it, which perhaps explains my instinct to post the answer that I did. In short: you're right, this is an overly cautious answer. –  Zev Chonoles May 27 '11 at 7:50