A little problem about monotonous functions

Last entry was about the fundamental theorem of algebra, a big result at the time (beginning of the 19th century), although it is no longer considered difficult. Today a would just like to talk about a little problem a friend came up with. Actually it occurred while he was teaching first year students. They had to solve an exercise, one of them asked a question, he reformulated it, and finally he obtained the following question.

Consider the vector space over the field \mathbb{R} of the functions from \mathbb{R} to \mathbb{R} (we will call it \mathcal{F}(\mathbb{R},\mathbb{R})). What are the subspaces of this vector space that contain only monotonous functions?

As he pointed out, there are three rather obvious examples:
-the null space (duh!),
-the line spanned by a monotonous function,
-the two-dimensional vector space spanned by a continuous function and a constant function (say 1).
In fact those are the only possibilities. That is rather easy to prove, but a bit long. Maybe it can be shortened. Anyway, we first need to rule out the case of three-dimensional vector spaces and higher.

Lemma 1

Let X be a set and \mathbb{K} a (commutative) field. Let \mathcal{F}(X, \mathbb{K}) be the vector space over \mathbb{K} of the functions from X to \mathbb{K}. The functions f_1,\dots,f_n taken in \mathcal{F}(X,\mathbb{K}) are linearly independent if, and only if, there exist x_1,\dots,x_n \in X such that

\displaystyle   \left|  \begin{array}{ccc}   f_1(x_1)&\cdots&f_n(x_1)\\   \vdots&\ddots&\vdots\\  f_1(x_n)&\vdots&f_n(x_n)   \end{array}   \right|   \neq 0.

Proof. One of the implication is simpler. If we are given f_1,\dots,f_n such that there is a non-zero determinant as stated, then f_1,\dots f_n are linearly independent. Indeed, if \alpha_1 f_1+\dots+\alpha_n f_n \equiv 0, then we have in particular

\displaystyle   \left\{   \begin{array}{lcl}   \alpha_1f_1(x_1)+\cdots+\alpha_n f_n(x_1)&=&0;\\   &\vdots&\\   \alpha_1 f_1(x_n)+\cdots+\alpha_n f_n(x_n)&=&0.   \end{array}   \right.

Because its determinant is not zero, the system in \alpha_1,\dots,\alpha_n has a unique solution, namely \alpha_1=\dots =\alpha_n=0.
The other implication is a rather straightforward induction on n. Assume that the implication holds for n \in \mathbb{N}^*. Let f_1,\dots , f_n, f_{n+1} be linearly independent functions. By the induction hypothesis, there exist x_1, \dots , x_n \in \mathbb{R} such that

\displaystyle   \left|   \begin{array}{ccc}   f_1(x_1)&\cdots&f_n(x_n)\\   \vdots   &\ddots&\vdots  \\   f_1(x_n)&\cdots&f_n(x_n)   \end{array}   \right|   \neq 0.

If, for all x \in \mathbb{R},

\displaystyle   \left|   \begin{array}{cccc}   f_1(x_1)&\cdots&f_n(x_1)&f_{n+1}(x_1)\\   \vdots&\ddots&\vdots&\vdots\\   f_1(x_n)&\cdots&f_n(x_n)&f_{n+1}(x_n)\\   f_1(x)&\cdots&f_n(x)&f_{n+1}(x)   \end{array}   \right|   =0.

then by developping the determinant along the last column, we find a_1, \dots ,a_n \in \mathbb{R} such that for all x,

\displaystyle   f_{n+1}(x)=a_1f_1(x)+\cdots +f_n(x).
This obviously contradicts the assumption that the function f_1, \dots , f_n, f_{n+1} are linearly independent.Therefore, there must be a x_{n+1} \in \mathbb{R} such that the determinant is not zero.

As a consequence of this lemma, we see that a subspace of the vector space \mathcal{F}(\mathbb{R},\mathbb{R}) that contains only monotonous functions is of dimension at most two. Indeed, if f, g, h \in \mathcal{F} are three linearly independent functions, then there exists x,y,z \in \mathbb{R} such that

\displaystyle   \left|   \begin{array}{lcl}   f(x)&g(x)&h(x)\\   f(y)&g(y)&h(y)\\   f(z)&g(z)&h(z)   \end{array}   \right|   \neq 0.

Then for all a, b, c \in \mathbb{R}, there exists \alpha, \beta, \gamma \in \mathbb{R} such that

\displaystyle   \left\{   \begin{array}{lcl}   \alpha f(x) +\beta g(x) + \gamma h(x)&=&a;\\   \alpha f(y) +\beta g(y) + \gamma h(y)&=&b;\\   \alpha f(z) +\beta g(z) + \gamma h(z)&=&c.   \end{array}   \right.

By choosing appropriate values for a, b and c, we can make the function \alpha f+ \beta g+\gamma h non monotonous.

We are halfway there. What we have to check now is that a two-dimensional subspace of \mathcal{F}(\mathbb{R},\mathbb{R}) that contains only monotonous functions must contain a constant function (that is not zero). I will give a proof by contraposition.

Lemma 2

Let f, g \in \mathcal{F}(\mathbb{R},\mathbb{R}) be two linearily independant functions such that none of the linear combinations
\alpha f+\beta g is constant. Then one of these combinations is non-monotonous.

Proof. By the previous lemma, there are x, y \in \mathbb{R} (we can assume that x<y) such that

\displaystyle   \left|   \begin{array}{cc}   f(x)&g(x)\\   f(y)&g(y)   \end{array}   \right|   \neq 0.

Let us chose a \neq 0. There exist \alpha, \beta \in \mathbb{R} such that

\displaystyle   \left\{   \begin{array}{lcl}   \alpha f(x)+\beta g(x)&=&a;\\   \alpha f(y)+\beta g(y)&=&a.   \end{array}   \right.

Since h=\alpha f+ \gamma g is not a constant, there is z \in \mathbb{R} such that f(z) \neq a. If x<z<y, then h is not monotonous. If z \notin (0,1), then by applying one, or both, of the transformations f(x) \mapsto -f(x) and f(x)\mapsto f(-x)
we only have to treat the case where x<y<z and h(z)>a, which is convenient. We will show that a small perturbation of h is non-monotonous.
We know that there exist \gamma, \delta \in \mathbb{R} such that

\displaystyle   \left\{   \begin{array}{lcl}   \gamma f(x)+\delta g(x)&=&1;\\   \gamma g(y)+\delta g(y)&=&-1.   \end{array}   \right.

If we set h_{\epsilon}=(\alpha+\epsilon\gamma)f+(\beta+\epsilon \delta)g, then h_{\epsilon}(x)=a+\epsilon, h_{\epsilon}(y)=a-\epsilon and h_{\epsilon}(z)=h(z)+\epsilon(\gamma f(z)+\delta g(z)). By choosing \epsilon>0 small enough, we get h_{\epsilon}(x)>h_{\epsilon}(y) and h_{\epsilon}(z)>h_{\epsilon}(x), and in that case h_{\epsilon} is not monotonous.

Not much of a result, but it was fun to do (not so fun to write it up).


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