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June 13

Homogeneous function, in the weak sense.

Is there any common name for the function $f$ which has a kind of "homogeneity" [of order 1] in the following weak sense:

There is $\alpha \ne 1,$ satisfying for every $x$ in the domain of the function: $f(\alpha x)=\alpha f(x).$

I have no objection to stipulating that $\alpha$ is also non-zero (or even positive), if that makes it easier to find the common name I'm looking for.

In particular: Is there any common name for the function $f$ that has a sort of "homogeneity" [of order 1] in an even weaker sense:

For every $x$ in the domain of the function, there is $\alpha \ne 1$ satisfying: $f(\alpha x)=\alpha f(x).$

Again, I have no objection to stipulating that $\alpha$ is also non-zero (or even positive), if that makes it easier to find the common name.

2A06:C701:7469:F000:21B3:2273:921C:D9AC (talk) 10:48, 13 June 2023 (UTC)

For the first question, an applicable term is "linear map" (also known as "linear function", but that term is ambiguous), assuming that the domain and codomain, which apparently admit scalar multiplication, are vector spaces. The sets $ℝ$ and $ℂ$ can be seen as one-dimensional vector spaces over themselves as fields. The requirement of preservation of addition is satisfied, since:

$\xcancel{f(x+y)=(x+y)f(1)=xf(1)+yf(1)=f(x)+f(y).}$

 --Lambiam 13:32, 13 June 2023 (UTC)

Unfortunately, here we can only guarantee $f(x+y)=(x+y)f(1)$ if $x+y=\alpha$, and we can also only guarantee $xf(1)=f(x)$ and $yf(1)=f(y)$ if $x=\alpha$ or $y=\alpha$ respectively. GalacticShoe (talk) 13:40, 13 June 2023 (UTC)

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Haven't found anything regarding naming conventions, but there are some interesting properties that one could consider for these kinds of functions. Note that for both notions of weak homogeneity, $f(0)=0$.

When a function $f$ is "weakly" homogeneous of order 1 for some parameter $\alpha >1$, it is completely defined by its values over $(-\alpha ,-1]\cup [1,\alpha )$, since every real number $x\ne 0$ can be written canonically and uniquely as ${\alpha }^{k}y$ for some $y\in (-\alpha ,-1]\cup [1,\alpha )$ and some $k\in \mathbb{Z}$, making $f(x)=f({\alpha }^{k}y)={\alpha }^{k}f(y)$. In fact, any values of $f$ over $(-\alpha ,-1]\cup [1,\alpha )$ work, no matter how pathological. The case of $0<\alpha <1$ is equivalent but with $(-\frac{1}{\alpha },-1]\cup [1,\frac{1}{\alpha })$, since $\frac{1}{\alpha }>1$. $\alpha =0$ yields a trivial function where $f(0)=0$ and the rest of $f$ is literally any function you want. For $\alpha <0$, however, it becomes more complicated.

When a function $f$ is "weakerly" homogeneous of order 1 meanwhile, intuitively what this means is that any line passing through the origin intersects with the graph of $f$ either not at all, or at least twice. If $f$ is differentiable then we can think of this as there being no tangent points for lines passing through the origin, but naturally $f$ in this case doesn't need to be differentiable, or even any particularly form of "nice"; any odd function, no matter how pathological, works. GalacticShoe (talk) 14:19, 13 June 2023 (UTC)

For the second question, the exponential function restricted to the positive real numbers has this property. Define function $g$ on the positive reals by $g(1)=1,g(a)=\frac{\log a}{a-1}(a\ne 1).$ Let $h$ be its inverse function, also defined on the positive reals, so $g(a)=b$ iff $a=h(b).$ Now, for given $x,$ take $\alpha =h(x).$ Then $\exp (\alpha x)=\alpha \exp (x),$ since

$\exp (\alpha x)=$ $\exp ((\alpha -1)x+x)=$ $\exp ((\alpha -1)g(\alpha )+x)=$ $\exp ((\log \alpha )+x)=$ $\alpha \exp (x).$

 --Lambiam 14:24, 13 June 2023 (UTC)

One minor caveat here; we also have to discount $x=1$, since it yields $\alpha =1$, which is disallowed. From a graph standpoint, this is because the line $y=ex$ intersects the graph of $y=e^x$ at a tangent point at $(1,e)$. GalacticShoe (talk) 14:35, 13 June 2023 (UTC)

Let $g$ be a periodic function with period $P.$ Define $f$ on the positive reals by

$f(x)=x\exp g(\log x).$

This function $f$ satisfies the first condition. For take $\alpha =\exp P.$ Then

$f(\alpha x)=$ $\alpha x\exp g(\log (\alpha x))=$ $\alpha x\exp g(\log \alpha +\log x)=$ $\alpha x\exp g(P+\log x)=$ $\alpha x\exp g(\log x)=$ $\alpha f(x).$

--Lambiam 16:12, 13 June 2023 (UTC)

In your last example, you could replace exp by any other function (including the identity function or any constant function), and still receive a weak homogeneity (in the first meaning). 2A06:C701:7469:F000:21B3:2273:921C:D9AC (talk) 18:43, 13 June 2023 (UTC)

Correct, the proof does not use any property of $\exp .$ Also, for any function $h$ it is replaced by, the function $g^{\prime }=h\circ g$ is periodic if $g$ is, so an extra function application after $g$ does not increase the generality.  --Lambiam 19:40, 13 June 2023 (UTC)

In fact, it can be seen that for functions on the positive reals, $f(x)=xg(\log x),$ for a periodic function $g,$ is the most general form. For suppose some function $f$ satisfies the identity $f(\alpha x)=\alpha f(x)$ for some positive $\alpha \ne 1.$ Define function $g$ by

$g(z)=\frac{f(\exp z)}{\exp z},$

and put $P=\log \alpha .$ Then

$g(z+P)=\frac{f(\exp (z+P))}{\exp (z+P)}=$ $\frac{f(\alpha \exp z)}{\alpha \exp z}=$ $\frac{\alpha f(\exp z)}{\alpha \exp z}=$ $\frac{f(\exp z)}{\exp z}=g(z).$

So $g$ is a periodic function, and the identity $g(z)=\frac{f(\exp z)}{\exp z}$ implies $f(x)=xg(\log x).$

--Lambiam 20:07, 13 June 2023 (UTC)

Note that this extends to the negative real numbers as well, given a possibly separate periodic $g$ of the same period as the original, with:

$$g(z)=\frac{f(-\exp z)}{-\exp z}$$

and $P=\log \alpha$ yielding:

$$g(z-P)=\frac{f(-\exp (z-P))}{-\exp (z-P)}=\frac{f(\frac{1}{a}*-\exp z)}{\frac{1}{a}*-\exp z}=\frac{\frac{1}{a}f(-\exp z)}{\frac{1}{a}*-\exp z}=\frac{f(-\exp z)}{-\exp z}=g(z)$$

and $f(x)=xg(\log -x)$.

This means that weak homogeneous functions in general are defined piecewise as:

$$f(x)=\{\begin{array}{cc}x{g}_1(\log x),& \text{if }x>0\\ 0,& \text{if }x=0\\ x{g}_2(\log -x),& \text{if }x<0\end{array}$$

where $g_1$ and $g_2$ are both periodic with period $\log \alpha$. GalacticShoe (talk) 22:41, 13 June 2023 (UTC)

In both meanings of weak homogeneity? 2A06:C701:7469:F000:21B3:2273:921C:D9AC (talk) 07:02, 14 June 2023 (UTC)

This is not the most general form for the even weaker sense with $\mathrm{\forall }x\mathrm{\exists }\alpha .$ The exponential function restricted to the positive reals minus $\{1\}$ has a sort of homogeneity in the even weaker sense, but cannot be expressed in this form, since $g(z)=\exp (\exp z-z)$ is not periodic.  --Lambiam 07:37, 14 June 2023 (UTC)

Homogeneous functiona are a subject of study btw. --RDBury (talk) 01:47, 14 June 2023 (UTC)

This thread is about weak homogeneity. 2A06:C701:7469:F000:21B3:2273:921C:D9AC (talk) 07:03, 14 June 2023 (UTC)

For homogeneity in the even weaker sense, let $h$ be a continuous function that is unbounded in either direction (positive and negative) while the set of its zeros is also unbounded. For example, it might be the function $h(u)=u\sin u,$ whose graph keeps making wider and wider swings. Now define:

$g(u)=h(u)+u,$

$f(x)=\exp (g(\log x)).$

Any function $f$ thus defined, whose domain is the set of positive reals, meets the definition. To show this, we need to establish that given $x>0$ there exists a value $\alpha \ne 1$ such that the equation $f(\alpha x)=\alpha f(x)$ is satisfied. In the following, we use $\lambda$ as shorthand for $\log x$ and $P$ as shorthand for $\log \alpha .$ Working out both sides of the equation, we find:

$f(\alpha x)=\exp (g(P+\lambda )),$

$\alpha f(x)=\exp (P+g(\lambda )).$

The equation has a solution iff the equation $g(P+\lambda )=P+g(\lambda )$ has a solution for $P\ne 0.$ After replacing $g$ by its definiens, simplification results in the equation

$h(P+\lambda )=h(\lambda ).$

With some handwaving (here the lecturer makes wider and wider up-and-down swings with their hand): the unboundedness conditions on function $h$ imply the existence of values $\kappa$ and $\mu ,$ both on the same side of $\lambda$, such that $h(\kappa )<h(\lambda )<h(\mu ).$ By the continuity of $h,$ there exists ${\lambda }^{\prime }\in (\min (\kappa ,\mu ),\max (\kappa ,\mu ))$ such that $h({\lambda }^{\prime })=h(\lambda ).$ Taking $P={\lambda }^{\prime }-\lambda$ gives us a solution.  --Lambiam 09:28, 14 June 2023 (UTC)

This is not the most general form of functions meeting the even weaker version. For example, if $p$ is any continuous non-constant periodic function and $q$ is any continuous function whose range is $ℝ,$ taking $h=p\circ q$ also gives us a function for which the equation $h({\lambda }^{\prime })=h(\lambda ),$ given $\lambda ,$ is guaranteed to have a solution for ${\lambda }^{\prime }\ne \lambda .$  --Lambiam 09:59, 14 June 2023 (UTC)

When $x\ne 0$, the condition that $\mathrm{\exists }\alpha \ne 1:f(\alpha x)=\alpha f(x)$ is equivalent to saying that $\mathrm{\exists }y=\alpha x\in ℝ:f(y)=\frac{y}{x}f(x)$, which is equivalent to $\mathrm{\exists }y\in ℝ:\frac{f(y)}{y}=\frac{f(x)}{x}$. When $x=0$, any $\alpha$ works as $f(\alpha x)=f(0)=0=\alpha f(x)$. As such, I'm pretty sure that the most general form of a function of weaker homogeneity is that a function $f$ is weaker-homogeneous if and only if $\frac{f(x)}{x}$ is nowhere-injective outside of $0$, and is $0$ at $0$. GalacticShoe (talk) 16:10, 14 June 2023 (UTC)

Note that weak homogeneous functions, by this definition, are immediately weaker homogeneous, as $\frac{f(x)}{x}$ is $0$ at $0$, is $g_1(\log x)$ for periodic $g_1$ for $x>0$ and thus nowhere injective on $x>0$, and is $g_2(\log -x)$ for periodic $g_2$ for $x<0$ and thus nowhere injective on $x<0$. GalacticShoe (talk) 16:13, 14 June 2023 (UTC)

As an example, the earlier demonstration of $\exp x$ on ${ℝ}^{+}-\{e\}$ being weaker-homogeneous results immediately from the fact that $\frac{\exp x}{x}$ is nowhere injective on the domain. GalacticShoe (talk) 16:17, 14 June 2023 (UTC)

I was just about to post the observation that the two cases I posted above have in common that they use a function $h$ with the strong anti-injectivity property that for any $u$ in its domain there exists $u^{\prime }\ne u$ such that $h(u^{\prime })=h(u).$ Defining $\phi (x)=\frac{f(x)}{x},$ function $h$ can be expressed as

$h=\log \circ \phi \circ \exp ,$

which immediately relates the anti-injectivity properties of $\phi$ and $h.$  --Lambiam 16:27, 14 June 2023 (UTC)