Wikipedia:Reference desk/Archives/Mathematics/2023 June 18

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June 18

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What anything, more specific can as simple/general as possible, should we know about a given function  , if for concluding that   is known to be a one-to-one correspondence?

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I have no objection to stipulating that   is from the set of positive numbers to iteself. Want to assume also that   is monotonic? No ojection. Continuous? No objection. Differntiable? No objection. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 04:42, 18 June 2023 (UTC)[reply]

If   is one-to-one and continuous on the positive reals, it is strictly monotonic. And a continuous strictly monotonic function is 1-1. So any function   that can be written as   for a strictly monotonic   will work. —Kusma (talk) 08:49, 18 June 2023 (UTC)[reply]
And under the assumption that   is continuous, this is a most general form for  .  --Lambiam 09:11, 18 June 2023 (UTC)[reply]
  where   is (continuous and) strictly monotonic can be defined precisely as   which is (continuous and) strictly monotonic in one direction over  , (continuous and) strictly monotonic in the other direction over  , and   at  . GalacticShoe (talk) 13:03, 18 June 2023 (UTC)[reply]
In general, an injection   is a bijection  , where   is uncountable. Moreover, if   and   is bijective, then naturally   is injective. So the most general form is   where   is bijective from   to some uncountable subset of  . Of course, this is a very general (and redundant) form and doesn't seem to have any nice properties, but that's at least partially because there's not a lot of structure to be gleaned from the one injectivity requirement. There are some very pathological examples one can construct here, mostly from bizarre  . Consider, for example, a bijection between the real numbers and the Cantor set. GalacticShoe (talk) 13:23, 18 June 2023 (UTC)[reply]
As an example of a bijection between the real numbers and the Cantor set, one can construct a bijection   defined by  , and then one can construct a bijection   defined by taking  , writing out its binary expansion, converting all  s to  s, then converting it from trinary to decimal, finally yielding the function   as a bijection  , which looks like this. The function that results from   looks like this, and does seem rather pathological for a function which, divided by  , is injective. For one, it looks like there should be quite a few lines that intersect the function in more than one place, but that's just a consequence of both the graphics used (Matlab draws the curve as a continuous line), as well as the gaps in the Cantor set. GalacticShoe (talk) 14:15, 18 June 2023 (UTC)[reply]

OP's clarification: Sorry for my mistake, which I've just fixed on the header: I meant just the opposite: For a given function   I would like to conclude that   is one-to-one, by a previous simple information about   (that can be srtictly monotonic or continuous/differentiable or defined from the set of positive numbers to itself, as you wish), without g being mentioned in that previous information about   2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:16, 18 June 2023 (UTC)[reply]

Note that the previous conversations apply pretty much entirely the other way too, so   being equal to   for   continuous and strictly monotonic implies   being continuous and injective, and   being equal to   for bijective   for   implies   being injective. GalacticShoe (talk) 13:32, 18 June 2023 (UTC)[reply]
A sufficient condition is to assume that   is differentiable and   is everywhere positive or everywhere negative. For example, any   such that   and   for every  . —Kusma (talk) 13:36, 18 June 2023 (UTC)[reply]
Quick note that the condition on   for every   should be   for   and   for  ; otherwise, you can get, for example,  ,  , and   which is not everywhere positive/negative. GalacticShoe (talk) 14:28, 18 June 2023 (UTC)[reply]
I am working only in positive numbers. As we are talking about a function written as  , we can't use   anyway. —Kusma (talk) 15:59, 18 June 2023 (UTC)[reply]
Fair point given that we are working with   specifically, I just thought it worth pointing out that the distinction is necessary in general. GalacticShoe (talk) 17:28, 18 June 2023 (UTC)[reply]

Starting anew

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Sorry for my mistakes in the previous thread. I thought, I had to strike out some words on the header, and to add other words, but now I realize my previous thought was a second mistake, so I decide to start anew, being as exact as possible, this time:

What I know is the following: Both   and   are continuous (and even differentiable) injections, defined for every positive number.

[Later addition: Hence their injectivity can be replaced by their strict monotonicity].   is also known to be strictly monotonic.

Besides this information, can I infer something else (anything non obvious) about   (rather than about   2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 16:30, 18 June 2023 (UTC)[reply]

Any continuous injective function is strictly monotonic. (For a proof, see Proposition 5.7.2 here; I only saw the if direction on Wikipedia.) So the strict monotonicity of   already follows from its given continuity and   is also known to be strictly monotonic.  --Lambiam 07:32, 19 June 2023 (UTC)[reply]
Thanks. I've just added your information [in brackets] to my original question on this thread. See the third paragraph above. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk)
If both   is continuous, then   is automatically continuous as well. In terms of   being monotonic, however, it is a necessary (but not sufficient) condition that   have no zeroes. If  , then   is continuous and either always positive or always negative except for at  , which immediately implies that all neighborhoods of   breaks the injectivity constraint. If   for   meanwhile, then   is   at   and  , meaning that all combined neighborhoods of   and   break the injectivity constraint. GalacticShoe (talk) 15:58, 19 June 2023 (UTC)[reply]
Beyond   being either all-positive or all-negative, there's actually not much one can glean when   is just continuous. For example, although the monotone convergence theorem guarantees that   has a limit in at least one direction, adding conditions on the limit doesn't seem to make   any more or less likely to be monotonic. For example, since   is not monotonic for all  , the function   has arbitrary limit  , yet yields   that is not monotonic. I have yet to consider if differentiability allows for some better properties, but I'm not particularly hopeful on that front, given that monotonic functions in general already are differentiable everywhere except a set of measure zero. GalacticShoe (talk) 16:30, 19 June 2023 (UTC)[reply]
OP's response. I think I have gleaned some details about   yet they are too specific, whereas I'm looking for more general (non-obvious) ones.
As mentioned above, both   and   are continuous (and even differentiable) injections. So here are some specific details I can glean about  
1.   defined over the set of real numbres, cannot be the specific function   for any real   because even though   is a continuous injection as required,   is not.
2.   defined over the set of real numbres, cannot be the specific function   for any odd positive exponential   because even though   is a continuous injection as required,   is not.
3.   defined over the set of positive numbres, cannot be the specific function   for any positive base   because even though   is a continuous injection as required,   is not.
4.   defined over the interval   cannot be the specific function   because even though   is a continuous injection as required,   is not.
So, could you fill in the blanks?
5.   cannot be any function having the following general (non-obvious) property: ____________________ , because even though   is a continuous injection as required,   is not.
Note that by "non-obvious" property, I intend to exclude any property which   is obviously not permitted to have, like: "  for some continuous (and even differentiable) function   that is not injective while   is".
2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 19:37, 19 June 2023 (UTC)[reply]
If   [defined over the set of positive numbres (OP's addition)] is differentiable, the graph of its derivative   cannot cross the positive x-axis, that is, the equation   has no positive solutions in  . But the equation   can also not have any positive solutions. Some examples. For   we have   everywhere. For   we have   For   we have   So none of these are OK. But for   we have   and for   we have   so these are fine. Also, for   we have   while   so   is OK provided that    --Lambiam 21:25, 19 June 2023 (UTC)[reply]
First, thank you for your response. Second, please notice I've inserted some words [in brackets] into your first sentence (after I addded some new paragraphs to my previous response), I hope it's ok (if it's not feel free to delete what I inserted into your first sentence). Third, you claim: "the equation   can also not have any positive solutions". But how can your claim fill in the blanks, in my paragraph #5? Some words should be inserted there, so what are they, according to your claim? 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 22:39, 19 June 2023 (UTC)[reply]
"it allows a (positive) solution of the equation  ". This is equivalent to: "the tangent to the graph of   for some   passes through the origin".  --Lambiam 22:51, 19 June 2023 (UTC)[reply]
Do you mean, that every continuous function   having a (positive?) solution for the equation   is an injection which satisfies that the function   is not an injection? Or you mean that every continuous injection   having a (positive?) solution for the equation   satisfies that the function   is not an injection? Or you mean something different? 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 01:39, 20 June 2023 (UTC)[reply]
A solution of the equation, if it exists, is a value of   satisfying the equation. It is positive if   The phrase was meant to fit in the slot of your #5, as requested. The assumption is that   is a continuous (and even differentiable) injection defined on   So,
  cannot be any continuous (and even differentiable) injection defined on   having the following general (non-obvious) property: it allows a (positive) solution of the equation   because even though   is a continuous injection as required,   is not.
--Lambiam 07:57, 20 June 2023 (UTC)[reply]
Thank you again, ever so much, for filling in the slot of my #5. Just to be sure, which one of the following sentences is necessarily true? Both?
  • Strong version: Every differentiable function   defined on  , and allowing a positive solution of the equation   is an injection which satisfies that the function   is not an injection.
  • Weak version: Every differentiable injection   defined on  , and allowing a positive solution of the equation   satisfies that the function   is not an injection.
Additionaly, what about the opposite direction of your claim: For every differentiable injection   defined on  , and satisfying that   is not injective, is it true to claim that the equation   has a (positive) solution?
2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:24, 20 June 2023 (UTC)[reply]
The strong version is false. Counterexample: take   Then   yet   I think both the weak version and its converse are valid: for any differentiable injection   defined on   function   is not injective if and only if   has a solution in    --Lambiam 22:24, 20 June 2023 (UTC)[reply]
Thanx ever so much. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 08:51, 21 June 2023 (UTC)[reply]
I'm sorry, but I have to retract part of this statement. Let   Both   and   are injective on   but   for   It appears that only the converse of the weak version is left standing.  --Lambiam 15:18, 21 June 2023 (UTC)[reply]
Now you deserve a double thanx, because of your honesty. Honesty is the best policy. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 21:02, 21 June 2023 (UTC)[reply]