Functions $f(x)$ for which the set of functions $af(x+b)$ is closed under addition

Question

Let $f(x)$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$. Consider the set of functions $af(x+b)$ for $a,b \in \mathbb{R}$. If that set is closed under addition, what can $f$ be?

I suspect the possible $f$'s are exponentials, sinusoids, constants and their products. (For example, the sum of any two sinusoids with the same period, each arbitrarily shifted and scaled, is another sinusoid with the same period - and I couldn't find any other periodic function with that property - that was my original motivation for the problem.) But I don't know how to prove there are no others.

Another funny note is that $f(x)=x$ yields a set which is almost closed under addition, except its additive closure also includes nonzero constant functions, which don't have the form $a(x+b)$.

Answer 1

So, what we have want is a description of those continuous $f: \mathbb{R} \to \mathbb{R}$ such that the the image of the function $(a,b) \mapsto af(x + b)$, where $a, b \in \mathbb{R}$, is closed under addition. We'll call this function $T:\mathbb{R}^2 \mapsto C(\mathbb{R})$ ie $T(a,b) = x \mapsto af(x + b)$.

Notice that the image of the plane under $T$ will also be closed under scalar multiplication, so its a vector subspace of $C(\mathbb{R})$.

If $f$ is the $0$ function, the image is just the $0$ function. In every other case, it has a subgroup that is isomorphic to $\mathbb{R}$, generated by $\{(a,0)\}$.

If that's all there is, then I think there's a pretty clear argument that we have a function of the form $x \mapsto A\cdot B^x$ where $a, b \in \mathbb{R}$ and $B>0$. To outline, the image of $\{(0,b)\}$ must be contained in the image of $\{(a,0)\}$. So, let $A= f(0)$ and $B = f(1)/f(0)$ ($f(0)$ being non zero is implied by not being in the boring case). I claim that $f(x) = A\cdot B^x$. To get there you need a lemma like "$f(x + b) = Cf(x)$ iff $f(x + 2b) = C^2f(x)$" and continuity.

I feel like there can only be two dimensions in the image, but I don't have a good proof of that, and I don't have a good argument treating that case, but I thought the treatment of the 1D case was worth putting in black and white.

Answer 2

This will only be a partial answer aiming to give a strong necessary condition in the case where $f$ is assumed to be uniformly continuous over $\mathbb{R}$, making use of the theory of tempered distributions and the fact that the Fourier transform acts "nicely" on delay equations. We will show that uniformly continuous solutions are necessarily of the form $x \mapsto \sum_{j = 1}^r \lambda_j e^{ip_j x}$, which covers all the solutions cited by OP.
This is probably not a sufficient condition (coupled with $f(\mathbb{R}) \subset \mathbb{R}$, ça va sans dire, as I would say in French), but I'll let someone else take over unless I have an idea for the more general case where uniform continuity is not assumed since it's a (relatively) harsh condition to impose.

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Assume for the rest of this answer that $f$ is uniformly continuous and non-constant (trivial case). Since $f$ goes from $\mathbb{R}$ to $\mathbb{R}$, we can find a modulus of continuity with finite values everywhere (usually you would have to allow infinite values), aka a function $\omega : [0, +\infty) \to [0, +\infty)$ continuous at $0$ such that $\omega(0) = 0$ and satisfying: $$\forall (x,y) \in \mathbb{R}^2,\quad |f(y) - f(x)| \leq \omega(|y - x|)$$

Now, for $h > 0$, let $(a_h, b_h) \in \mathbb{R}^2$ such that $f(\cdot + h) - f(\cdot) =: a_h f(\cdot + b_h)$. This pair exists by OP's assumption on $f$. Note also that $a_h = 0$ is equivalent to $f$ being $h$-periodic.
Then, we get, for all $x \in \mathbb{R}$ and $h > 0$: $$|a_h f(x)| = |a_h f(x - b_h + b_h)| = |f(x - b_h + h) - f(x - b_h)| \leq \omega(h)$$ The immediate effects of this inequality are twofold:

• Since $f \neq 0$, we get $|a_h| \leq \frac{\omega(h)}{|f(x_0)|}$ for some $x_0$ such that $f(x_0) \neq 0$, hence $a_h$ tends to $0$ as $h$ tends to $0^+$;
• Assume by contradiction that $f$ is not bounded. Then there exists a sequence $(x_n)_n \subset \mathbb{R}$ such that $(|f(x_n)|)_n$ diverges to $+\infty$.
  However that would mean that we'll have the following for all $h > 0$: $$0 \leq |a_h| \leq \frac{\omega(h)}{|f(x_n)|} \xrightarrow[n \to +\infty]{} 0$$ Yet $a_h$ is constant in regard to $n$, hence we'd obtain $a_h = 0$ for all $h > 0$, which is impossible as $f$ would be continuous and $1$- and $\pi$-periodic for example thus constant, and in particular bounded, absurd.
  Therefore a uniformly continuous non-constant $f$ satisfying the desired condition is bounded, and there exists at least one $h_0$ for which $a_{h_0} \neq 0$.

Now we can get to the distribution aspect. Since $f$ is bounded and continuous, it induces a tempered distribution, and we can thus consider its Fourier transform in the distributional sense.
Taking the Fourier transform in the equation $a_h f(x + b_h) + f(x) - f(x + h) = 0$ yields: $$\forall h > 0, \forall \xi \in \mathbb{R},\quad \left(a_h e^{ib_h\xi} + 1 - e^{ih\xi}\right) \hat{f}(\xi) = 0$$ The support of $\hat{f}$ is thus contained in the set $E := \left\{\xi \in \mathbb{R} \mid \forall h > 0,\, a_h e^{ib_h\xi} + 1 - e^{ih\xi} = 0\right\}$. Let's show that this latter set is discrete, countable and bounded, thus finite.
If we allow $\xi$ to take complex values, then it's clear that $g: \xi \in \mathbb{C} \mapsto a_{h_0} e^{ib_{h_0}\xi} + 1 - e^{ih_0\xi}$ is an entire function of $\xi$ distinct from $0$, as for $\xi = 0$ we find $g(0) = a_{h_0} \neq 0$, hence by the identity theorem for holomorphic functions the set of zeroes $Z_g$ of $g$ is countable and has no accumulation points in $\mathbb{C}$.
But $E \subset Z_g$, hence $E$ is also countable and discrete. It only remains to see that $E$ is bounded.

Because $a_h \to 0$ when $h \to 0^+$, there exists $\eta$ such that, for all $0 < h \leq \eta$, we have: $|a_h| \leq \frac{1}{3}$.
By the second triangle inequality, this provides the following: $$\forall h \in (0, \eta], \forall \xi \in \mathbb{R},\quad \left|a_h e^{ib_h\xi} + 1 - e^{ih\xi}\right| \geq \left|1 - e^{ih\xi}\right| - \left|a_h e^{ib_h\xi}\right| \overset{b_h \in \mathbb{R}}{\geq} \left| 1 - e^{ih\xi}\right| - \frac{1}{3}$$ This implies that for $\xi$s belonging to $F_h := \left\{\xi \in \mathbb{R} \mid \left|1 - e^{ih\xi}\right| \geq \frac{1}{2}\right\}$, we'll have: $$\left|a_h e^{ib_h\xi} + 1 - e^{ih\xi}\right| \geq \left| 1 - e^{ih\xi}\right| - \frac{1}{3} \geq \frac{1}{2} - \frac{1}{3} > 0$$ In other words: $$\forall h \in (0,\eta],\quad F_h \subset \mathbb{R} \setminus E$$ which means that $E$ will be bounded if the union of the $F_h$s contains $\mathbb{R} \setminus I$ for some bounded interval $I$.
This comes down to looking for $\xi$s such that $\left|1 - e^{ih\xi}\right|^2 \geq \frac{1}{4}$, yet: $$\left|1 - e^{ih\xi}\right|^2 = (1 - \cos(h\xi))^2 + \sin^2(h\xi) = 2(1 - \cos(h\xi))$$ thus $\xi \in F_h$ iff $1 - \cos(h\xi) \geq \frac{1}{8}$, which is itself equivalent to $\cos(h\xi) \leq \frac{7}{8}$ (you can probably pick a different constant instead of the $\frac{1}{2}$ from earlier so you can have something nicer than $\frac{7}{8}$ here but oh well).
This is fulfilled iff $\xi \in \left[\frac{\arccos\left(\frac{7}{8}\right)}{h}, \frac{2\pi - \arccos\left(\frac{7}{8}\right)}{h}\right] + \frac{2\pi}{h}\mathbb{Z}$, and so in particular: $$\left[-\frac{2\pi - \arccos\left(\frac{7}{8}\right)}{h}, -\frac{\arccos\left(\frac{7}{8}\right)}{h}\right] \cup \left[\frac{\arccos\left(\frac{7}{8}\right)}{h}, \frac{2\pi - \arccos\left(\frac{7}{8}\right)}{h}\right] \subset F_h$$ By choosing strategic $h$s such that the intervals above overlap, which is possible since they vary "continuously" in respect to $h$ but I'll leave the details to the reader, we can conclude that: $$\mathbb{R} \setminus \left(\frac{\arccos\left(\frac{7}{8}\right)}{\eta}, \frac{2\pi - \arccos\left(\frac{7}{8}\right)}{\eta}\right) =: \mathbb{R} \subset I \subset \bigcup_{0 < h \leq \eta} F_h \subset \mathbb{R} \setminus E$$ which finally gives: $E \subset I$, and $E$ is bounded.

$E$ is countable, discrete and bounded, thus finite, therefore so is the support of $\hat{f}$ since it was contained in $E$.
However, we have a characterisation of the distributions whose Fourier transforms have finite support: they are exactly the functions $x \mapsto \sum_{j = 1}^r \lambda_j x^{k_j} e^{ip_j x}$ for some $r \in \mathbb{N}$, some $(k_1, \dots, k_r) \in \mathbb{N}^r$, some $(\lambda_1, \dots, \lambda_r) \in (\mathbb{C} \setminus \{0\})^r$ and some $(p_1, \dots, p_r) \in \mathbb{R}^r$ (see my topical question linked here: Distributions whose Fourier transforms have finite support. It's possible that I'm wrong so feel free to correct me on the linked post if so, especially since it affects this answer too!). Thus: $$ f : x \mapsto \sum_{j = 1}^r \lambda_j x^{k_j} e^{ip_j x}$$

Moreover, our $f$ is bounded due to our previous observations. Looking at the expression given, this is not possible if any of the $k_j$s is non-zero (details left to the reader. One way to proceed is by setting apart the contributions to the leading monomial with the second triangle inequality), hence we actually get: $$ f : x \mapsto \sum_{j = 1}^r \lambda_j e^{ip_j x}$$

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EDIT $15/02/2024$: No chance yet with other more general functions, but let me reduce the number of possible solutions in the uniformly continuous case by showing that with the condition $f(\mathbb{R}) \subset \mathbb{R}$ the $\lambda_j$s can be chosen real and the exponentials can be turned into cosines. I don't know if I'll search for exactly which of these sums of cosines is a solution, given that it might be more calculatory than anything else (question mark?) but at least the set of solution-candidates will actually be real-valued this time...

Consider $f$ of the form $f : x \mapsto \sum_{j = 1}^r \lambda_j e^{ip_j x}$ exactly as described above and with $r \geq 1$ so that it is non-constant. We can assume that the $p_j$s are pairwise distinct, up to adding terms with the same $e^{ip_j \cdot}$ together. Finally, suppose that $f(\mathbb{R}) \subset \mathbb{R}$.

All the derivatives of $f$ at $0$ will be real due to being able to the difference-quotients lying in $\mathbb{R}$ at each step, yet we can also obtain these equalities simply by differentiating the function as usual: $$\forall k \in \mathbb{N},\quad \mathbb{R} \owns f^{(k)}(0) = \sum_{j = 1}^r \lambda_j p_j^k$$ therefore: $$\forall k \in \mathbb{N},\quad \operatorname{Im}\left(\sum_{j = 1}^r \lambda_j p_j^k\right) = \sum_{j = 1}^r \operatorname{Im}(\lambda_j) p_j^k = 0$$

In particular, this holds for $k \in \{0, \dots, r - 1\}$, hence: $$\pmatrix{1 & 1 &\cdots & 1\\ p_1 & p_2 & \cdots & p_r \\ p_1^2 & p_2^2 & \cdots & p_r^2\\ \vdots & \vdots & {} & \vdots\\ \\ p_1^{r-1} & p_2^{r-1} & \cdots & p_r^{r-1}} \pmatrix{\operatorname{Im}(\lambda_1) \\ \vdots \\ \operatorname{Im}(\lambda_r)} = 0$$ The Vandermonde matrix on the left is invertible by assumption, thus $\pmatrix{\operatorname{Im}(\lambda_1) \\ \vdots \\ \operatorname{Im}(\lambda_r)} = 0$, and all the $\lambda_j$s are necessarily real.

Now, $f$ being real-valued is equivalent to $f = \bar{f}$, yet $\overline{f(x)} = \sum_{j = 1}^r \lambda_j e^{-ip_j x}$, thus: $$f(x) = \frac{1}{2}\left(f(x) + \overline{f(x)}\right) = \frac{1}{2}\left(\sum_{j=1}^r \lambda_j e^{ip_j x} + \lambda_j e^{-ip_j x}\right) = \sum_{j = 1}^r \lambda_j \cos(p_j x)$$

which is the desired result for this section.

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Due to how uniform continuity was used at the very start to get a convergence condition on $(a_h)_h$, I do not know if it's possible to reuse the same logic in the general case, however do note that we really only needed that $|a_h| \leq c < 1$ for $h \leq \eta_0$ for some $\eta_0$, which implies that there should be at least a more general case.
In the other direction, one should find the same result if $|a_h| \geq d > 1$.

In all cases, and as long as $f$ induces a tempered distribution, the argument using the identity theorem works the same, hence $f$ has a Fourier transform with discrete and countable support, which is already a big limitation by itself.

Note also that only the $b_h$s being real mattered, if $f$ were complex-valued and the $a_h$s were allowed to be complex whereas the $b_h$s were imposed to stay real the reasoning would still hold.

Finally, periodic continuous functions are uniformly continuous and thus are all covered here.