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June 18

What anything, more specific can as simple/general as possible, should we know about a given function $f$, if for concluding that $\frac{f(x)}{x}$ is known to be a one-to-one correspondence?

I have no objection to stipulating that $f$ is from the set of positive numbers to iteself. Want to assume also that $f$ 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)

If $g:x\mapsto f(x)/x$ 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 $f$ that can be written as $f(x)=g(x)x$ for a strictly monotonic $g$ will work. —Kusma (talk) 08:49, 18 June 2023 (UTC)

And under the assumption that $f$ is continuous, this is a most general form for $f$.  --Lambiam 09:11, 18 June 2023 (UTC)

$f(x)=g(x)x$ where $g$ is (continuous and) strictly monotonic can be defined precisely as $f$ which is (continuous and) strictly monotonic in one direction over ${ℝ}^{-}$, (continuous and) strictly monotonic in the other direction over ${ℝ}^{+}$, and $0$ at $0$. GalacticShoe (talk) 13:03, 18 June 2023 (UTC)

In general, an injection $g:ℝ\to ℝ$ is a bijection $g:ℝ\to g(ℝ)$, where $g(ℝ)\subseteq ℝ$ is uncountable. Moreover, if $D\subseteq ℝ$ and $g:ℝ\to D$ is bijective, then naturally $g:ℝ\to ℝ$ is injective. So the most general form is $f(x)=xg(x)$ where $g$ is bijective from $ℝ$ to some uncountable subset of $ℝ$. 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 $g:ℝ\to D$. Consider, for example, a bijection between the real numbers and the Cantor set. GalacticShoe (talk) 13:23, 18 June 2023 (UTC)

As an example of a bijection between the real numbers and the Cantor set, one can construct a bijection $f:ℝ\to (0,1)$ defined by $f(x)=1/(1+\exp (-x))$, and then one can construct a bijection $g:(0,1)\to 𝒞$ defined by taking $x\in (0,1)$, writing out its binary expansion, converting all $1$s to $2$s, then converting it from trinary to decimal, finally yielding the function $h=g\circ f$ as a bijection $h:ℝ\to 𝒞$, which looks like this. The function that results from $xh(x)$ looks like this, and does seem rather pathological for a function which, divided by $x$, 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)

OP's clarification: Sorry for my mistake, which I've just fixed on the header: I meant just the opposite: For a given function $f,$ I would like to conclude that $g:x\mapsto f(x)/x$ is one-to-one, by a previous simple information about $f$ (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 $f.$ 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:16, 18 June 2023 (UTC)

Note that the previous conversations apply pretty much entirely the other way too, so $f(x)$ being equal to $xg(x)$ for $g$ continuous and strictly monotonic implies $f$ being continuous and injective, and $f(x)$ being equal to $xg(x)$ for bijective $g:ℝ\to D$ for $D\subseteq ℝ$ implies $f$ being injective. GalacticShoe (talk) 13:32, 18 June 2023 (UTC)

A sufficient condition is to assume that $f$ is differentiable and $x{f}^{\prime }(x)-f(x)$ is everywhere positive or everywhere negative. For example, any $f$ such that $f(x)>0$ and $f^{\prime }(x)\le 0$ for every $x$. —Kusma (talk) 13:36, 18 June 2023 (UTC)

Quick note that the condition on $f^{\prime }(x)\le 0$ for every $x$ should be $f^{\prime }(x)\le 0$ for $x\ge 0$ and $f^{\prime }(x)\ge 0$ for $x<0$; otherwise, you can get, for example, $f(x)=\exp (-x)$, $f^{\prime }(x)=-\exp (-x)$, and $x{f}^{\prime }(x)-f(x)=-x\exp (-x)-\exp (-x)$ which is not everywhere positive/negative. GalacticShoe (talk) 14:28, 18 June 2023 (UTC)

I am working only in positive numbers. As we are talking about a function written as $f(x)/x$, we can't use $x=0$ anyway. —Kusma (talk) 15:59, 18 June 2023 (UTC)

Fair point given that we are working with $f(x)/x$ specifically, I just thought it worth pointing out that the distinction is necessary in general. GalacticShoe (talk) 17:28, 18 June 2023 (UTC)

Starting anew

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 $g$ and $f:x\mapsto xg(x)$ are continuous (and even differentiable) injections, defined for every positive number.

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

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

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 $f(x)$ already follows from its given continuity and $g$ is also known to be strictly monotonic.  --Lambiam 07:32, 19 June 2023 (UTC)

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 $g(x)$ is continuous, then $xg(x)$ is automatically continuous as well. In terms of $g(x)$ being monotonic, however, it is a necessary (but not sufficient) condition that $g(x)$ have no zeroes. If $g(0)=0$, then $xg(x)$ is continuous and either always positive or always negative except for at $0$, which immediately implies that all neighborhoods of $0$ breaks the injectivity constraint. If $g(x_0)=0$ for $x_0\ne 0$ meanwhile, then $xg(x)$ is $0$ at $0$ and $x_0$, meaning that all combined neighborhoods of $0$ and $x_0$ break the injectivity constraint. GalacticShoe (talk) 15:58, 19 June 2023 (UTC)

Beyond $g$ being either all-positive or all-negative, there's actually not much one can glean when $g$ is just continuous. For example, although the monotone convergence theorem guarantees that $g$ has a limit in at least one direction, adding conditions on the limit doesn't seem to make $xg(x)$ any more or less likely to be monotonic. For example, since $x(e^x+N)$ is not monotonic for all $0\le N<1/e^2\approx 0.1353$, the function $g(x)=10K{e}^x+K$ has arbitrary limit $K\ne 0$, yet yields $xg(x)$ 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)

OP's response. I think I have gleaned some details about $f,$ yet they are too specific, whereas I'm looking for more general (non-obvious) ones.

As mentioned above, both $g$ and $f:x\mapsto xg(x)$ are continuous (and even differentiable) injections. So here are some specific details I can glean about $f:$

1. $f,$ defined over the set of real numbres, cannot be the specific function $f(x)=Bx,$ for any real $B\ne 0,$ because even though $f$ is a continuous injection as required, $g$ is not.

2. $f,$ defined over the set of real numbres, cannot be the specific function $f(x)=x^B,$ for any odd positive exponential $B,$ because even though $f$ is a continuous injection as required, $g$ is not.

3. $f,$ defined over the set of positive numbres, cannot be the specific function $f(x)={\log }_B(x),$ for any positive base $B\ne 1,$ because even though $f$ is a continuous injection as required, $g$ is not.

4. $f,$ defined over the interval $(0,1),$ cannot be the specific function $f(x)=\arcsin \sqrt{x},$ because even though $f$ is a continuous injection as required, $g$ is not.

So, could you fill in the blanks?

5. $f$ cannot be any function having the following general (non-obvious) property: ____________________ , because even though $f$ is a continuous injection as required, $g$ is not.

Note that by "non-obvious" property, I intend to exclude any property which $f$ is obviously not permitted to have, like: "$f(x)=xg(x)$ for some continuous (and even differentiable) function $g$ that is not injective while $f$ is".

2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 19:37, 19 June 2023 (UTC)

If $f$ [defined over the set of positive numbres (OP's addition)] is differentiable, the graph of its derivative $f^{\prime }$ cannot cross the positive x-axis, that is, the equation $f^{\prime }(x)=0$ has no positive solutions in $x$. But the equation $f^{\prime }(x)=g(x)$ can also not have any positive solutions. Some examples. For $f(x)=Bx$ we have $f^{\prime }(x)=g(x)$ everywhere. For $f(x)={\log }_B(x),B>0,B\ne 1,$ we have $f^{\prime }(e)=g(e)=e^{-1}{\log }_{B}e.$ For $f(x)=B^x,B>1,$ we have $f^{\prime }({\log }_{B}e)=g({\log }_{B}e)=e\log B.$ So none of these are OK. But for $f(x)=x^B,B\ne 0,B\ne 1,$ we have $f^{\prime }(x)=B{x}^{B-1}\ne g(x)=x^{B-1},$ and for $f(x)=x\exp x,$ we have $f^{\prime }(x)=(x+1)\exp x\ne g(x)=\exp x,$ so these are fine. Also, for $f(x)=B^x,0<B<1,$ we have $f^{\prime }(x)=(\log B)B^x<0,$ while $g(x)>0,$ so $f(x)=B^x$ is OK provided that $0<B<1.$  --Lambiam 21:25, 19 June 2023 (UTC)

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 $f^{\prime }(x)=g(x)$ 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)

"it allows a (positive) solution of the equation $f^{\prime }(x)=f(x)/x$". This is equivalent to: "the tangent to the graph of $y=f(x),$ for some $x>0,$ passes through the origin".  --Lambiam 22:51, 19 June 2023 (UTC)

Do you mean, that every continuous function $f,$ having a (positive?) solution for the equation $f^{\prime }(x)=f(x)/x,$ is an injection which satisfies that the function $f(x)/x$ is not an injection? Or you mean that every continuous injection $f,$ having a (positive?) solution for the equation $f^{\prime }(x)=f(x)/x,$ satisfies that the function $f(x)/x$ is not an injection? Or you mean something different? 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 01:39, 20 June 2023 (UTC)

A solution of the equation, if it exists, is a value of $x$ satisfying the equation. It is positive if $x>0.$ The phrase was meant to fit in the slot of your #5, as requested. The assumption is that $f$ is a continuous (and even differentiable) injection defined on ${ℝ}^{+}.$ So,

$f$ cannot be any continuous (and even differentiable) injection defined on ${ℝ}^{+}$ having the following general (non-obvious) property: it allows a (positive) solution of the equation $f^{\prime }(x)=f(x)/x,$ because even though $f$ is a continuous injection as required, $g$ is not.

--Lambiam 07:57, 20 June 2023 (UTC)

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 $f,$ defined on ${ℝ}^{+}$, and allowing a positive solution of the equation $f^{\prime }(x)=f(x)/x,$ is an injection which satisfies that the function $f(x)/x$ is not an injection.
• Weak version: Every differentiable injection $f,$ defined on ${ℝ}^{+}$, and allowing a positive solution of the equation $f^{\prime }(x)=f(x)/x,$ satisfies that the function $f(x)/x$ is not an injection.

Additionaly, what about the opposite direction of your claim: For every differentiable injection $f,$ defined on ${ℝ}^{+}$, and satisfying that $f(x)/x$ is not injective, is it true to claim that the equation $f^{\prime }(x)=f(x)/x$ has a (positive) solution?

2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:24, 20 June 2023 (UTC)

The strong version is false. Counterexample: take $f(x)=x^2-4x+9.$ Then $f^{\prime }(3)=f(3)/3=2,$ yet $f(1)=f(3)=6.$ I think both the weak version and its converse are valid: for any differentiable injection $f$ defined on ${ℝ}^{+},$ function $x\mapsto f(x)/x$ is not injective if and only if $f^{\prime }(x)=f(x)/x$ has a solution in ${ℝ}^{+}.$  --Lambiam 22:24, 20 June 2023 (UTC)

Thanx ever so much. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 08:51, 21 June 2023 (UTC)

I'm sorry, but I have to retract part of this statement. Let $f(x)=x^2+x\sin x.$ Both $f$ and $x\mapsto f(x)/x$ are injective on ${ℝ}^{+},$ but $f^{\prime }(x)=f(x)/x$ for $x=2k\pi ,k\in {\mathbb{Z}}^{+}.$ It appears that only the converse of the weak version is left standing.  --Lambiam 15:18, 21 June 2023 (UTC)

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)