# Are separable, continuous, monotonic and scale invariant real-valued functions everywhere differentiable?

Consider a function $$f:\mathbb{R}_+^2\rightarrow\mathbb{R}$$ of two non-negative real variables (or more generally of several real variables) that is increasing in each argument, continuous, additively (or multiplicatively) separable, that is, it can be written in the form $$f(x,y)=a(x)+b(y)$$ for functions of one variable, $$a$$ and $$b$$, and satisfies the following notion of "scale invariance": For each $$\lambda>0$$ and $$(x,y),(\tilde{x},\tilde{y})\in\mathbb{R}_+^2$$ we have $$f(x,y)=f(\tilde{x},\tilde{y})\Leftrightarrow f(\lambda x,\lambda y)=f(\lambda \tilde{x},\lambda \tilde{y}).$$ Is there an example of such a function that is not everywhere differentiable? We tried hard to find such an example. For example, we know that if such functions exist, they have to be non-differentiable on a dense null set, by the scale invariance property.

Edit: The original question was asked for non-continuous functions and it was answered negatively. In the edited version we ask the question for continuous functions.

• I updated my answer to cover the continuous case, too. By the way, how did you come up with this functional equation? Oct 21, 2022 at 23:24

If we allow $$f$$ to be discontinuous, then the answer is yes: $$f$$ need not be differentiable.

We can choose $$a(x)$$ and $$b(y)$$ so that their ranges $$A$$ and $$B$$ are Cantor-like sets which have the following property: $$\text{if \alpha, \alpha' \in A and \beta, \beta' \in B, and \alpha + \beta = \alpha' + \beta', then \alpha = \alpha' and \beta = \beta'.}$$ Then $$f(x, y) = f(x', y')$$ implies $$x = x'$$ and $$y = y'$$: just use the above property with $$\alpha = a(x)$$, $$\beta = b(y)$$, $$\alpha' = a(x')$$, $$\beta' = b(y')$$. This means that the scale-invariance property is trivially satisfied.

Examples of such sets $$A$$ and $$B$$ are easy to construct using decimal expansions. For instance, $$A$$ can be the set of real numbers which can be written using only $$0$$ and $$1$$, while $$B$$ — with $$0$$ and $$2$$. Then the function $$a(x)$$ can be defined as follows: in order to compute $$a(x)$$, write the binary expansion of $$x$$ and interpret it as the decimal expansion of $$a(x)$$. The function $$b(y)$$ can be defined in a similar way, or simply as $$b(y) = 2 a(y)$$.

On the other hand, if we require $$f$$ to be continuous, then the answer is no: $$f$$ is necessarily differentiable.

Step 1. Fix a positive real $$\lambda \geqslant 1$$. Since $$f(1, 1) \leqslant f(\lambda, 1) \leqslant f(\lambda, \lambda)$$, there is a number $$\mu = \mu(\lambda) \in [1, \lambda]$$ such that $$f(\mu, \mu) = f(\lambda, 1)$$ (here we use continuity of $$f$$). By scale-invariance, with $$\lambda = x / y$$, we have $$f(x, y) = f(\lambda y, y) = f(\mu y, \mu y)$$. A similar argument works if $$0 < \lambda \leqslant 1$$. Thus, if we denote $$\phi(x) = f(x, x)$$, then $$f(x, y) = \phi(y \mu(\tfrac xy)) .$$

Step 2. If $$f(x, y) = a(x) + b(y)$$, then the above equality takes form $$a(x) + b(y) = \phi(y \mu(\tfrac xy)) .$$ In particular, if $$x_1 < x_2$$, we have $$a(x_2) - a(x_1) = (a(x_2) + b(x_1)) - (a(x_1) + b(x_1)) = \phi(x_1 \mu(\tfrac{x_2}{x_1})) - \phi(x_1)$$ and $$a(x_2) - a(x_1) = (a(x_2) + b(x_2)) - (a(x_1) + b(x_2)) = \phi(x_2) - \phi(x_2 \mu(\tfrac{x_1}{x_2})) .$$ If $$x_1 = t$$ and $$x_2 = s t$$, then the above identities read $$a(s t) - a(t) = \phi(t \mu(s)) - \phi(t) = \phi(s t) - \phi(\nu(s) t),$$ where $$\nu(\lambda) = \lambda \mu(1/\lambda)$$ also lies between $$1$$ and $$\lambda$$. Similarly, $$b(s t) - b(t) = \phi(s t) - \phi(t \mu(s)) = \phi(\nu(s) t) - \phi(t) .$$

Step 3. If $$\mu(s) = 1$$ for some $$s \ne 1$$, then $$a(s t) - a(t) = 0$$, and hence $$a$$ is constant. Similarly, if $$\mu(s) = s$$ for some $$s \ne 1$$, then $$\nu(1/s) = 1$$, so that $$b(t / s) - b(t) = 0$$ and hence $$b$$ is constant. Thus, in what follows we assume that $$\mu(s)$$ lies strictly between $$1$$ and $$s$$ for all $$s \ne 1$$.

Step 4. We already know that $$\phi(t \mu(s)) - \phi(t) = \phi(s t) - \phi(\nu(s) t) .$$ If $$\psi$$ is a nonnegative, smooth, compactly supported function and $$\Phi(t) = \int_0^\infty \phi(u t) \psi(u) dt ,$$ then $$\Phi$$ is smooth and $$\Phi(t \mu(s)) - \Phi(t) = \Phi(s t) - \Phi(\nu(s) t) .$$ We consider the Taylor expansion of the above expressions near $$s = 1$$. Unfortunately, this gets rather technical (I guess a simpler approach is possible, but I fail to see one straight away).

First, we find that $$\lim_{s \to 1} \frac{s - \nu(s)}{\mu(s) - 1} = \frac{\lim_{s \to 1} \frac{\Phi(t \mu(s)) - \Phi(t)}{\mu(s) - 1}}{\lim_{s \to 1} \frac{\Phi(s t) - \Phi(\nu(s) t)}{s - \nu(s)}} = \frac{t \Phi'(t)}{t \Phi'(t)} = 1 ,$$ and similarly $$\lim_{s \to 1} \frac{s - \mu(s)}{\nu(s) - 1} = 1 .$$ Therefore, \begin{aligned} & \lim_{s \to 1} \frac{(s - 1)^2 - (\nu(s) - 1)^2 - (\mu(s) - 1)^2}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad = \lim_{s \to 1} \frac{(s - \mu(s)) (s - \nu(s))}{(\mu(s) - 1)(\nu(s) - 1)} + \lim_{s \to 1} \frac{(s - \mu(s)) (\mu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad \qquad + \lim_{s \to 1} \frac{(s - \nu(s)) (\nu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} - \lim_{s \to 1} \frac{(\mu(s) - 1)(\nu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad = 1 + 1 + 1 - 1 = 2 . \end{aligned} It follows that $$0 = \lim_{s \to 1} \frac{\Phi(s t) - \Phi(\nu(s) t) - \Phi(t \mu(s)) + \Phi(t)}{(\mu(s) - 1) (\nu(s) - 1)} = p t \Phi'(t) + t^2 \Phi''(t) ,$$ where a finite limit $$p = \lim_{s \to 1} \frac{s - \nu(s) - \mu(s) + 1}{(\mu(s) - 1) (\nu(s) - 1)}$$ necessarily exists. We conclude that $$\Phi(t) = c_1 t^{1 - p} + c_2$$, unless $$p = 1$$, in which case $$\Phi(t) = c_1 \log t + c_2$$.

Since the function $$\psi$$ was arbitrary, we necessarily have $$\phi(t) = C_1 t^{1 - p} + C_2$$, unless $$p = 1$$, in which case $$\Phi(t) = C_1 \log t + C_2$$.

Step 5. Recall that $$a(s t) - a(t) = \phi(t \mu(s)) - \phi(t) .$$ Dividing both sides by $$(\mu(s) - 1)$$ and passing to the limit as $$s \to 1$$, we find that $$\lim_{s \to 1} \frac{a(s t) - a(t)}{\mu(s) - 1} = t \phi'(t) .$$ Since this holds for every $$t$$, the function $$a$$ is necessarily differentiable, with $$\frac{t a'(t)}{\mu'(0)} = t \phi'(t) .$$ Thus, $$a(t) = C_3 + \mu'(0) \phi(t)$$. Similarly, $$b(t) = C_4 + \nu'(0) \phi(t)$$.

Summary. We have shown that $$a(x)$$ and $$b(y)$$ are differentiable, and in fact for some constants $$c_1, c_2, c_3, c_4, \gamma$$, with $$c_2 \gamma, c_4 \gamma > 0$$ if $$\gamma \ne 0$$, we have $$a(x) = c_1 + c_2 x^\gamma, \qquad b(y) = c_3 + c_4 y^\gamma ,$$ except for the case $$\gamma = 0$$, where we have $$c_2, c_4 > 0$$ and $$a(x) = c_1 + c_2 \log x, \qquad b(y) = c_3 + c_4 \log y .$$

• There must be a simpler solution in the continuous case, but at this moment I can only give the above rather technical argument. Oct 21, 2022 at 23:23