Is there a non-constant continous function $f:\mathbb{R}\rightarrow \mathbb{R}$ and matrices $A=\begin{pmatrix} a_1 & 0\\ 0 & a_2\\ \end{pmatrix}$ and $B=\begin{pmatrix} b_1 & 0\\ 0 & b_2\\ \end{pmatrix}$ for which there does not exist any matrices $C\in \mathrm{Mat}_{d\times 2},D \in \mathrm{Mat}_{2\times d}$ and vectors $c \in \mathbb{R}^d,e\in \mathbb{R}^2$ such that:

  • $ D f_d\left(C \begin{pmatrix} x\\ x \end{pmatrix} +c \right) + e = Af_2 \left(B \begin{pmatrix} x\\ x \end{pmatrix} \right) \qquad (\forall x \in \mathbb{R}) $

  • $D,e,C,$ and $c$ only have non-zero entries,

where $f_i(x)= (f(x_1),\dots,f(x_n))$.

  • $\begingroup$ Since $C(x , x)^t = xC(1, 1)^t$, that is $x$ times the sum of the two colums of $C$, we may replace $C$ with a fully unconstrained $d$-dimensional vector $c'$. $\endgroup$
    – Luc Guyot
    Feb 12, 2020 at 17:44
  • $\begingroup$ Don't $f(x) = \text{exp}(x)$ and $A = B = Id_2$ fit the bill? $\endgroup$
    – Luc Guyot
    Feb 12, 2020 at 18:34
  • $\begingroup$ @LucGuyot Is this only because $exp(x)$ is non-affine and injective? $\endgroup$
    – ABIM
    Feb 12, 2020 at 19:04
  • $\begingroup$ No, the function $x \mapsto x^3$ does not fit your requirements. $\endgroup$
    – Luc Guyot
    Feb 12, 2020 at 22:55
  • 1
    $\begingroup$ Assuming that you allow $C(1, 1)^t$ to be zero, it is equivalent to ask whether there is a non-zero continuous function $f$ which doesn't lie in the linear span of $\{x \mapsto f(cx + c')\}_{c \in \mathbb{R}, c' \in \mathbb{R} \setminus \{0\}}$.Therefore $x \mapsto \text{exp}(x)$ is not an example; I retract my suggestion. But $x \mapsto \text{exp}(-1 /x^2)$ does the job. $\endgroup$
    – Luc Guyot
    Feb 13, 2020 at 1:29

1 Answer 1


Yes, we can find such a triple $(f, A, B)$.

Let us first observe that OP's question can be rephrased as follows.

Question. Find a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f$ does not lie in the $\mathbb{R}$-linear span $L(f)$ of $\{ x \mapsto f(cx + c') \}_{(c, c') \in \mathbb{R} \times \mathbb{R} \setminus \{0\}}$.

For instance, the continuous extension $f$ of $x \mapsto \text{exp}(- 1 / x^2)$ is such that $f \notin L(f)$. Indeed, any function in $L(f) $ is analytic in a neighbourhood of 0 whereas $f$ isn't. I believe that many analytic functions, including $f(x) = \text{exp}(x^2)$, can be shown to satisfy $f \notin L(f)$, but a simple proof of this fact still eludes me.

By contrast, if $f$ is a real-valued polynomial function over $\mathbb{R}$ then $L(f)$ is the $\mathbb{R}$-vector space of the polynomial functions of degree at most $\deg(f)$, so that $f \in L(f)$. To see this, one may use the Taylor series of $f(x + c)$ together with a well-known result on Vandermonde matrices.

It is also immediate to check that periodic functions and the exponential function $f(x) = \text{exp}(x)$ satisfy $f \in L(f)$. They can be used to build algebras of functions satisfying this property.


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