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I proved an inequality, say $|f(x)| \leq C |g(x)|$. I am sure that $C$ is a number between $3.4$ and $59.8$. However I don't want to write like this, or, I don't want just say $|f(x)| \leq 60 |g(x)|$, since there are a lot of inequalities like this in my article and the range of $C$ may vary from one to one (but still in this manner).

In this case, is it Ok (without any confusion) to write $|f(x)| \leq C |g(x)|$, for some absolute constant $C$? What is the best way to write this?

I ask this question since in Terence Tao's book, he wrote sth. like "Here C denotes various absolute constants depending only on n, G, and N.". So Terence Tao considered "an absolute constant" is a constant depending on some other fixed objects, like the dimension of domains, the order of a group,...

Hana
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  • Not sure what you are after. Of course you can just write $|f(x)|≤C|g(x)|$ and just declare that $C$ is a constant depending only on $f,g$ but if you do that then you have lost the information regarding the magnitude of $C$. – lulu Aug 23 '21 at 00:25
  • @Lulu No, $C$ does not depend on $f$ and $g$. Here $C$ is some number (what we are uncertain) between 3.4 and 59.8.

    The main concern here is that: there are many like this in my article. Let's say, then we have |g(x)| < C |h(x)|, for some C between 345.12 and 456.45.

     – Hana Aug 23 '21 at 00:27
  • Of course $C$ depends on $f$ and $g$. There is no real number $C$ such that $|f(x)|≤C|g(x)|$ for all functions $f,g$. Or maybe you left off information...maybe $f,g$ belong to some collection with some special properties. Then $C$ could just depend on that collection. – lulu Aug 23 '21 at 00:28
  • @Lulu I think you didn't get my point. Here f and g are some functions satisfying certain properties, what we don't care here. – Hana Aug 23 '21 at 00:30
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    Your question is not clear. Where are the quantifiers? Does your inequality hold for all $f,g$? For all $x$? For some special choices of $f,g,x$? – lulu Aug 23 '21 at 00:31
  • So, let's say I proved the following very difficult inequality $(x+1)^2 \leq C (x^2 +1)$, for all real $x$, for some positive constant $C$, what I am sure, between 1.2 and 4.5. But I don't know exactly what should I give for $C$. In this case, is it ok to just say "an absolute constant". That's what I asked. – Hana Aug 23 '21 at 00:36
  • Again, that $C$ would obviously depend on the functions $(x+1)^2$ and $(x^2+1)$. But I see no advantage in discarding the known range, if there is one. So I'd say what you said. that there is a $C$ for which this holds (presumably for all $x$) and that $C$ is known to be in the given range. – lulu Aug 23 '21 at 00:38
  • What happens if there are 35 inequalities like this. I don't want to declare all ranges of $C$. – Hana Aug 23 '21 at 00:39
  • Then don't. I really don't understand, though. The range is important. Why throw that information away? But of course you are free to discard the information if you like. Maybe add a note saying that there are known ranges for the various constants. – lulu Aug 23 '21 at 00:42
  • So, my question points out only to "is it ok ..."? – Hana Aug 23 '21 at 00:44
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    It's up to you what information you put in. You can say " there is a constant $C$ such that $|f(x)|≤C|g(x)|$ for all $x$" or you could say "" there is a constant $C$ such that $|f(x)|≤C|g(x)|$ for all $x$ and $1<C<3$" or whatever range. – lulu Aug 23 '21 at 00:46
  • Is it okay according to whom? There's no universal style guide/rules set on this – Alan Aug 23 '21 at 02:25

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