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 of the functions from to (we will call it ). 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 ).
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.
Let be a set and a (commutative) field. Let be the vector space over of the functions from to . The functions taken in are linearly independent if, and only if, there exist such that
Proof. One of the implication is simpler. If we are given such that there is a non-zero determinant as stated, then are linearly independent. Indeed, if , then we have in particular
Because its determinant is not zero, the system in has a unique solution, namely .
The other implication is a rather straightforward induction on . Assume that the implication holds for . Let be linearly independent functions. By the induction hypothesis, there exist such that
If, for all ,
then by developping the determinant along the last column, we find such that for all ,
This obviously contradicts the assumption that the function are linearly independent.Therefore, there must be a such that the determinant is not zero.
As a consequence of this lemma, we see that a subspace of the vector space that contains only monotonous functions is of dimension at most two. Indeed, if are three linearly independent functions, then there exists such that
Then for all , there exists such that
By choosing appropriate values for , and , we can make the function non monotonous.
We are halfway there. What we have to check now is that a two-dimensional subspace of that contains only monotonous functions must contain a constant function (that is not zero). I will give a proof by contraposition.
Let be two linearily independant functions such that none of the linear combinations
is constant. Then one of these combinations is non-monotonous.
Proof. By the previous lemma, there are (we can assume that ) such that
Let us chose . There exist such that
Since is not a constant, there is such that . If , then is not monotonous. If , then by applying one, or both, of the transformations and
we only have to treat the case where and , which is convenient. We will show that a small perturbation of is non-monotonous.
We know that there exist such that
If we set , then , and . By choosing small enough, we get and , and in that case is not monotonous.
Not much of a result, but it was fun to do (not so fun to write it up).