Differentiable $f$ such that the set of translates of multiples of $f$ is a vector space of dimension two

Question

How can we derive all of the differentiable functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $V=\{af_b : a,b \in \mathbb{R}\}$ is a vector space of dimension two, where $f_b\colon \mathbb{R} \to \mathbb{R}$ is defined as the translate $f_b(x)=f(x+b)$?

An example is $f=\sin$. The set $V$ contains the linearly independent functions $\sin$ and $\cos$ (since $\cos(x)=\sin(x+\frac{\pi}{2})$). By elementary trigonometry we have $c\sin(x+\phi) = s\sin x + t\cos x$ for any $c, \phi \in \mathbb{R}$ and some $s,t \in \mathbb{R}$, so it follows that $V$ is a vector space of dimension two spanned by $\sin$ and $\cos$.

Answer 1

It would hold for the solution space of a second order ODE which is translation invariant on $\mathbb{R}$. There are a few out there: $$ \frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0,\;\;B, C \mbox{ are constants }. $$ Your example fits into this category. The solution space is 2-dimensional, and any translation of a solution is again a solution because of translation invariance; therefore, translates of solutions are linear combinations of any fixed basis of solutions. Linear sums of such solutions are also solutions.

Answer 2

Let $c \in \mathbb{R}$ be such that $\{f,f_c\}$ is a basis for $V$, where $f_c(x)=f(x+c)$. Then by definition there are unique functions $a,b\colon \mathbb{R} \to \mathbb{R}$ such that $f_t=a(t)f+b(t)f_c$ for all $t \in \mathbb{R}$. Now for any $x_1, x_2 \in \mathbb{R}$, let $$M(x_1,x_2)=\left[\begin{matrix} f(x_1) & f_c(x_1) \\ f(x_2) & f_c(x_2)\end{matrix}\right]$$ There must exist $x_1,x_2$ such that $M(x_1,x_2)$ is nonsingular, since otherwise $f$ would be an exponential function and $V$ would only have dimension one. Then for such $x_1$ and $x_3$, $\left[\begin{matrix} a(t) \\ b(t)\end{matrix}\right]=(M(x_1,x_2))^{-1}\left[\begin{matrix}f_t(x_1) \\ f_t(x_2)\end{matrix}\right]=(M(x_1,x_2))^{-1}\left[\begin{matrix}f_{x_1}(t) \\ f_{x_2}(t)\end{matrix}\right]$, so $a$ and $b$ can be written as linear combinations of $f_{x_1}$ and $f_{x_2}$. Since $f$ and its translates are differentiable, this implies that $a$ and $b$ are differentiable.

Now taking $\frac{\partial}{\partial t}$ of both sides of $f_t=a(t)f+b(t)f_c$ at $t=0$, we get $f'=a'(0)f+b'(0)f_c$. This is a linear combination of $f$ and $f_c$, so $f' \in V$. Note that $f$ and $f'$ are linearly independent (since otherwise $f$ would be an exponential and $V$ would have dimension one), so $\{f, f'\}$ span $V$. Since $f', f'_c \in V$, we also have $f'' = a'(0)f'+b'(0)f_c' \in V$, so $f'' \in \text{span}\{f, f''\}$. Therefore $f$ satisfies some ODE of the form

$$ \frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0 \:\:\:\:\:B,C\text{ constants } $$

As T.A.E. pointed out, the solutions to this ODE are all solutions to to the question (excluding $f(x)=c e^{rx}$, since in this case $V$ only has dimension one), so the solutions to the question are:

• $f(x)= c_1e^{r_1 x}+c_2 e^{r_2 x}\:\:\:$ ($c_1, c_2 \neq 0$ and $r_1 \neq r_2$)

• $f(x) = (c_1+c_2 x)e^{rx}\:\:\:\:\,\,$ ($c_2 \neq 0$)

• $f(x) = ce^{rx}\sin(sx+t)\,\,$ ($c, s \neq 0$)

(Postscript: Note that we did not need to initially assume that $V$ was a vector space, just that $V$ contains all $af_b$ for $a,b \in \mathbb{R}$ and is contained in $\text{span}\{f, f_c\}$ for some translate $f_c$ which is linearly independent of $f$. For then it still follows that $f', f'' \in \text{span}\{f, f_c\}$, so $f$ obeys the ODE. It is then an observed property of the solutions that $V$ is indeed a vector space.)