Log (uniform) continuous functions

Question

I am interested in functions $f: \mathbb R_+\to \mathbb R_+$ such that $\log \circ f \circ \exp$ is uniformly continuous. In other words \begin{align} \forall_{c}\, \exists_{c'}, \forall_{ x,x'\in[x/c',xc']}\, f(x')\in[f(x)/c, f(x)c]. \end{align} This definition includes functions such as $x\mapsto x^2$ and $x\mapsto x^{-1}$ which are not uniformly continuous.

I'm wondering what the right term for such functions is? I have also looked for "multiplicatively uniformly continuous", also with no luck.

If the range of $f$ is compact, can we say that $f^{-1}$ is also log uniformly continuous? Can we say that if $f$ and $g$ are Log uniform continuous, then so is $fg$?

Answer 1

I don't know if there's a standard term for your concept. (I'll use your term "multiplicatively uniformly continuous.") To answer your other questions:

If $f$ has a bounded range (assuming you mean bounded rather than compact), is $f^{-1}$ m.u.c.?

No. A counterexample is $f(x) = e^{-1/x}$, which has range $(0, 1)$. Thus $f^{-1}(x) = -\frac{1}{\log x}$ and $$\log \circ f^{-1} \circ \exp x = -\log (-x),$$ defined on the interval $(-\infty, 0)$. This is obviously not uniformly continuous.

Is the product of two m.u.c. functions also m.u.c.?

Yes. Let $f$ and $g$ be m.u.c. Then \begin{align*} \log \circ (x \mapsto f(x) g(x)) \circ \exp x &= \log \left[f(e^x)g(e^x)\right] \\ &= \log f(e^{x}) + \log g(e^x) \\ &= (\log \circ f \circ \exp + \log \circ g \circ \exp)(x) \end{align*} is the sum of two uniformly continuous functions and is therefore uniformly continuous itself.

Answer 2

I assume that $\Bbb R_+$ is the set of positive real numbers and $\log=\log_e$.

If the range of $f$ is compact, can we say that $f^{-1}$ is also log uniformly continuous?

If the function $f$ is log uniformly continuous then a function $1/f$ is log uniformly continuous too, because a function $\log\circ( 1/f)\circ \exp=-\log\circ f\circ \exp $ is uniformly continuous.

If a function $f^{-1}$ exists then the function $f$ is injective. If, moreover, the image of $f$ is a compact $K$ then $f^{-1}$ is not determined on the whole set $\Bbb R_+$, so formally it is not log uniformly continuous. Moreover, $f$ is not even continuous, because each continuous injective map from $\Bbb R_+$ to $\Bbb R$ is (strictly) monotonic, so its image is non-compact.

Can we say that if $f$ and $g$ are Log uniform continuous, then so is $fg$?

If both functions $f$ and $g$ are log uniformly continuous functions then their composition $fg$ is also log uniformly continuous, because a function $\log\circ f\circ g\circ\exp= \log\circ f \circ\exp\circ\log\circ g\circ \exp$ is uniformly continuous as a composition of uniformly continuous functions.

Answer 3

$\newcommand{\r}{\mathbb R_+}$$\newcommand{\a}[1]{\left(#1\right)}$ $\newcommand{\b}[1]{\left[#1\right]}$Let us prove an if and only if criteria for such functions. (Recall $\r = (0,\infty)$ )

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First, start with $f$ being luc, so for all $c > 1$ there is $c' > 1$ such that for all $x',x \in \r$ such that $\frac{x'}{x} \in \b{\frac 1{c'},c'}$ we have $\frac{f(x')}{f(x)} \in \b{\frac{1}{c},c}$.

Now, fix $y > x$, and $C > 1$. For this $C > 1$, get a $C' > 1$ satisfying the conditions of luc.

Now, we define $N \geq 0$ , $l \geq 0$ and the points $z_i$ as follows : $z_0 = x$, $z_1 = x C'^{l},...,z_k = x C'^{kl}$, and then find some $N$ such that $yC'^{-1} <z_{N} < y$.

That is, we want $xC'^{Nl}> yC'^{-1}$ so $Nl + 1 > \log_{C'} \a{\frac yx}$, and $y > xC'^{Nl}$ so $\log_{C'} \a{\frac yx}>Nl$. In short, $\frac{\log_{C'} \a{\frac yx}}l > N >\frac{\log_{C'}\a{\frac yx} - 1}{l}$, so with $l<\frac 12$ such $N$ can be found.

Note that $\frac{z_i}{z_{i-1}} = C'^{l} \in \b{\frac 1{C'},C'}$ for $i=1,...,N$, and $\frac{y}{z_N} \in \b{\frac 1{C'},C'}$. Therefore: $$ \frac {f(y)}{f(x)} = \prod_{i=1}^N \frac{f(z_i)}{f(z_{i-1})} \times \frac{f(y)}{f(z_N)} \leq C^{N+1} \times \frac yx \leq C\a{\frac yx}^{\frac 1l + 1} $$

Similarly : $$ \frac{f(y)}{f(x)} \geq C^{-(N+1)} \times \frac yx \geq C^{-1} \a{\frac yx}^{1-\frac 1l } $$

In conclusion we significantly weaken the above and obtain for $K = \frac 1l$: $$ C^{-1} \a{\frac yx}^{-K} \leq \frac{f(y)}{f(x)}\leq C \a{\frac yx}^{K} \tag{1} $$

We state this condition as : for all $C > 1$ there exists $2 < K < \infty$ such that $(1)$ holds for every $y > x$.

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Is the converse true? Suppose that $f$ is such that $g$ satisfies the given conditions. Then we want to find a $C'>1$ satisfying luc for given $C>1$. Now, $\sqrt C >1$, so note that by $(1)$ there is a $2 < L < \infty$ such that we have for every $y > x$ : $$ C^{-\frac 12}\a{\frac yx}^{-(L+1)} \leq \frac {f(y)}{f(x)} \leq \sqrt C \a{\frac yx}^{L+1} $$

Now, make $C^{-1} \leq C^{-\frac 12} \a{\frac yx}^{-(L+1)}$, which happens when $\frac yx \leq C^\frac 1{{2L+2}}$. By symmetry, we get that $C' = C^{\frac 1{2L+2}}$ works.

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We state this as the luc theorem!

$f : \r \to \r$ is luc if and only if for all $C > 1$ there is $2 < K < \infty$ such that for every $y \geq x$, $$\a{C \a{\frac yx}^{K}}^{-1} \leq \frac{f(y)}{f(x)} \leq C \a{\frac yx}^K$$

In words, the ratio $\frac{f(y)}{f(x)}$ can be controlled by a large enough power of $\frac yx$ to lie above certain arbitrarily given bounds (stricly) either side of $1$.

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How would we prove that the sum of luc functions is luc? Note that for any $y > x$ we have $\frac{f(y) + g(y)}{f(x) + g(x)}$ lies between $\frac{f(y)}{g(y)}$ and $\frac{f(x)}{g(x)}$. In other words, it is greater than the minimum of these and less than the maximum.

So, given $C > 1$, if $K_f$ and $K_g$ work for $f$ and $g$ then the maximum of $K_f,K_g$ works for the sum.

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The product of luc functions? It is clear from definition, but you just take $K_f + K_g$ here.

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The inverse of a function? The same $K$ works, just see the condition!

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The problem with subtraction is the domain must remain fixed. So I will explore what happens here.

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Noting that sums , products and inverses are luc, we may iterate and get the following result :

Given polynomials $p(x),q(x)$ which are positive on the positive real line, their quotient is luc.

Which generalizes your examples and provides a host of other examples.

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It is interesting to note that in the luc theorem, if the condition held for $C = 1$ then $\frac{f(y)}{f(x)}$ would be bounded either side by a fixed power (and its inverse) in $\frac yx$ (Take logarithms and see what this implies about $\log f$?). I urge you to find an luc function which does not satisfy this criteria with $C=1$, showing the strictness is important.

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