# Non-existent matrices with "essential zeros"

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

• 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 $c'$. Feb 12, 2020 at 17:44
• Don't $f(x) = \text{exp}(x)$ and $A = B = Id_2$ fit the bill? Feb 12, 2020 at 18:34
• @LucGuyot Is this only because $exp(x)$ is non-affine and injective?
– ABIM
Feb 12, 2020 at 19:04
• No, the function $x \mapsto x^3$ does not fit your requirements. Feb 12, 2020 at 22:55
• 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. Feb 13, 2020 at 1:29

## 1 Answer

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.