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

How can we derive all of the continuous 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 $\cos$.

This question has been answered in the case that $f$ is differentiable – $f$ is any solution of the ODE $f''+Bf'+Cf=0$ besides $ce^{rx}$ – but are there other non-differentiable continuous solutions to the problem, or are all the solutions differentiable?

All continuous solutions are differentiable.

Let $\mathcal{C}$ be the space of continuous functions over $\mathbb{R}$. Let $f \in \mathcal{C}$ be a function such that $V = \langle f_b : b \in \mathbb{R} \rangle$, the span of $f_b$ as a subspace of $\mathcal{C}$ is two dimenisonal. Let $$B = \bigg\{ b \in \mathbb{R} : f, f_b \text{ linear dependent on }\mathcal{C} \bigg\}$$ It is clear $B$ is a subgroup of $\mathbb{R}$ for addition. If $0 \in B$ is not an isolated point of $B$, then it is easy to see $B$ is dense in $\mathbb{R}$. Using continuity of $f_b$ w.r.t $b$, it is not hard to verify $f$ is an exponential function and leads to a contradiction that $V$ is one dimensional. As a result, $0$ is an isolated point of $B$.

WOLOG, we will assume $B \cap (-L,L) = \{ 0 \}$ for some $L > 1$ and take $f$ and $f_1$ as a basis of $V$. Let $A : \mathbb{R} \to M_{2}(\mathbb{R})$ be the $2 \times 2$ matrix valued function on $\mathbb{R}$ defined by:

\begin{align} \begin{pmatrix}f_{b}\\f_{b+1}\end{pmatrix} = A(b) \begin{pmatrix}f\\f_1\end{pmatrix} \end{align}\tag{*1} Since $f$ and $f_1$ is linear independent, there are $x, y \in \mathbb{R}$ such that the matrix $\begin{pmatrix}f(x) & f(y)\\f_1(x)&f_1(y)\end{pmatrix}$ is invertible. From $(*1)$, we have $$\begin{pmatrix}f_b(x)&f_b(y)\\f_{b+1}(x)&f_{b+1}(y)\end{pmatrix} = A(b) \begin{pmatrix}f(x) & f(y)\\f_1(x)&f_1(y)\end{pmatrix}$$

Since the LHS is a continuous function in $b$, so does $A(b)$. For any $b, c \in \mathbb{R}$, it is clear $A$ satisfies:

$$\begin{cases} & A(b) A(c) = A(b+c) = A(c)A(b)\\ \text{ and } & A(b) A(-b) = A(-b)A(b) = A(0) = I_2\tag{*2} \end{cases}$$

This implies $A(b)$ is always invertible.

Since $A(b)$ is continuous in $b$ and $A(0) = I_2$, there exists a $\epsilon > 0$ such that $\| A(b)-I_2 \| < \frac12$ for $b \in (-\epsilon, \epsilon)$. This implies

$$\left\| \int_0^b \left( A(t) - I_2 \right) dt \right\| \le \int_0^b \left\| A(t) - I_2 \right\| dy < \frac{b}{2}\quad\text{ for }b \in (0,\epsilon)$$

As a result, $\quad\displaystyle\int_0^b A(t) dt = b I_2 + \int_0^b \left(A(t) - I_2\right) dt\quad$ is invertible over $(0,\epsilon)$.

Let $D : (0,\epsilon) \to M_2(\mathbb{R})$ be the function $\quad\displaystyle D(b) = \left( A(b) - 1 \right)\left( \int_0^b A(t) dt \right)^{-1}$.
For any $b,c \in (0,\epsilon)$, we have

$$\int_0^{b+c} A(t) dt = \left( \int_0^b + \int_b^{b+c} \right) A(t) dt = I_2 \left( \int_0^b A(t) dt \right) + A(b) \left( \int_0^c A(t) dt \right)$$ Exchange the role of $b, c$ and subtract, we get \begin{align} & \left(A(b) - I_2\right)\left( \int_0^c A(t) dt \right) -\left(A(c) - I_2\right)\left( \int_0^b A(t) dt \right) = 0\\ \implies & \left(D(b) - D(c)\right) \left( \int_0^b A(t) dt \right) \left( \int_0^c A(t) dt \right) = 0\\ \implies & D(b) = D(c) \end{align}

i.e. $D(b)$ is a constant matrix over $(0,\epsilon)$. We will use the same $D$ to denote this constant matrix. Over $(0,\epsilon)$, $A(b)$ satisfies an integral equation

$$A(b) = I_2 + D \int_0^b A(t) dt$$ and hence $A(b) = e^{D b}$ there. Using $(*2)$, we see this expression is valid for all $b$. From this, we can conclude $A$ and hence $f$ is differentiable.

• Brilliant! Thank you for answering. – Malper Jan 1 '14 at 6:36