2

This might be quite trivial and I may be missing something obvious. Let $V$ be a vector space of functions from $\mathbb R$ into itself such that every function in $V$ is monotonic. Does it follow that the dimension of $V$ is at most $2$?

$\{ax+b: a,b \in \mathbb R\}$ shows that the dimension can be $2$.

Suppose $f$ and $g$ are linearly independent elements of $V$. By changing signs we may suppose that $f$ and $g$ are both increasing. If $f+ng$ is monotonically decreasing for infinitely many values of $n$ and $x<y$ then $f(x)+ng(x)\geq f(y)+ng(y)$ and $f(x)-f(y) \geq n (g(y)-g(x))$. Since $g(x) \leq g(y)$, this is a contradiction unless $g(x)=g(y)$. Thus, $g$ is a constant in this case.

I am unable to finish the proof and any help would be appreciated.

Rócherz
  • 3,976
Kavi Rama Murthy
  • 311,013
  • 2
    This might sound silly. But $e^{x}$ is monotonic and $e^{-x}$ is monotonic. But $cosh(x)$ is not. So does it even meet the criteria for vector space? . Please don't mind . I am just a student. – Mr.Gandalf Sauron Sep 28 '21 at 07:03
  • 1
    It is my assumption that all functions in $V$ are montonic. I am considering a very special type of vector space. @Mr.GandalfSauron – Kavi Rama Murthy Sep 28 '21 at 07:15
  • 3
    It's a good point @Mr.GandalfSauron, but I think Kavi is considering a vector space consisting only of monotone functions, not the set of all monotone functions. So, by your logic, if $V$ contains $e^x$, then it cannot contain $e^{-x}$ and vice-versa. – Theo Bendit Sep 28 '21 at 07:16
  • I think the problem might be even bigger - what is the zero vector in your vector space? Because the zero function is definitely not monotonic. – ConMan Sep 28 '21 at 07:28
  • 1
    @ConMan OP said that constants are monotone; so I would presume that the implied meaning is not ‘strictly’ monotone – Calvin Khor Sep 28 '21 at 07:30
  • 1
    Zero function is both monotonically increasing and monotonically decreasing. – Kavi Rama Murthy Sep 28 '21 at 07:30
  • Right, so weakly monotone, not strictly monotone. – ConMan Sep 28 '21 at 07:32

1 Answers1

4

We might as well add in the constant functions; $\operatorname{span}(V \cup \{1\})$ will also consist only of monotone functions. So, let's assume without loss of generality that the constant functions are indeed in $V$.

Suppose $V$ has dimension at least $3$ (possibly infinite). Then we may extend $\{1\}$ to a linearly independent set containing (at least) two other functions $f$ and $g$.

For any $a < b$, I claim there is a non-zero function in $\operatorname{span}\{f, g, 1\}$ that is $0$ on $\{a, b\}$. Consider the function $$(g(b) - g(a))(f - f(a)1) - (f(b) - f(a))(g - g(a)1) \in \operatorname{span}\{f, g, 1\}.$$ If the above function is the $0$ function, then the linear independence of $\{1, f, g\}$ implies $g(b) = g(a)$ and $f(b) = f(a)$. If one of the preceding numbers is $0$, then $f$ or $g$ will fill this role. Otherwise, $g(b)f - f(b)g$ will fill this role. So, either which way, we can find such a function, which we will call $h_{a, b}$.

Since $h_{a, b}$ is monotone, we therefore see that $h_{a, b}(x) = 0$ for all $x \in [a, b]$.

Now, fix $x_0 > 0$, and consider $h_{-x_0, x_0}$. Since $h_{-x_0, x_0} \neq 0$, there must exist some $x_1 > |x_0|$ such that $h_{-x_0, x_0}(x_1)$ or $h_{-x_0, x_0}(-x_1)$ is non-zero. Similarly, we can, for $n = 1, 2, 3$ find $x_n$ such that $h_{-x_{n-1}, x_{n-1}}(x_n)$ or $h_{-x_{n-1}, x_{n-1}}(-x_n)$ is non-zero.

Note that $\{h_{-x_n, x_n} : n = 0, 1, 2, 3\}$ is linearly independent, which we can see by restricting it to the set $\{\pm x_0, \pm x_1, \pm x_2, \pm x_3\}$. However, this is a subset of $\operatorname{span} \{f, g, 1\}$, which is of dimension at most $3$ (well, equal to $3$ by assumption). This is a contradiction, so $\{f, g, 1\}$ must be linearly dependent.

Theo Bendit
  • 50,900