Does there exist a 3-dimensional subspace of real functions consisting only of monotone functions? This is Exercise 1.O from the book 
Van Rooij, Schikhof: A Second Course on Real Functions.

The set of the monotone functions on $[0,1]$ contains all polynomial
  functions of degree $\le 1$. These form a two-dimensional vector space. Does the set of
  all monotone functions contain a three-dimensional vector space?

 A: $\newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\R}{\mathbb R}\newcommand{\intrv}[2]{[{#1},{#2}]}$
Lemma. Let $\Zobr{f,g}{\intrv01}{\R}$ be functions such that $f(0)=g(0)=0$ and the function $af(x)+bg(x)$ is monotone for any $a,b\in R$.
Then $f=0$ or $g=cf$ for some constant $c\in\R$.
Proof. Let $f\ne 0$. Let us denote the space consisting of all linear combinations of $f$ and $g$ by $V$. We assume that all functions in $V$ are monotone.
W.l.o.g we can assume that $f$ is non-decreasing. (Otherwise we can use the same proof for $-f$.)
Let us take an $x_0>0$ such that $f(x_0)>0$.
Put
\begin{align*}
c&:=\frac{g(x_0)}{f(x_0)}\\
h(x)&:=g(x)-cf(x)
\end{align*}
We have $h(0)=h(x_0)=0$, which implies $h(t)=0$ for every $t\in\intrv0{x_0}$. (Since $h$ is monotone.)
a) If $h=0$ then $g=cf$.
b) Suppose that $h\ne 0$. Then there exists a point $y_0$ such that $h(y_0)\ne 0$. We know that $y_0\notin\intrv0{x_0}$
This implies $0<x_0<y_0$. We have
\begin{align*}
0&=f(0)<f(x_0)\le f(y_0)\\
0&=h(0)=h(x_0)\ne h(y_0)
\end{align*}
W.l.o.g we may assume $h(y_0)>0$. (Otherwise we can work with $-h$.)
b.1) Suppose that $f(x_0)=f(y_0)$ and define $h_1=f-h$. Clearly $h_1\in V$, but
$$0<h_1(x_0)=f(x_0)>h_1(y_1)=f(x_0)-h(y_0),$$
so $h_1$ is not monotone.
b.2) Now suppose that $f(x_0)<f(y_0)$.
In this case we define
$$h_1:=f-2h\frac{f(y_0)-f(x_0)}{h(y_0)}.$$
Clearly $h_1\in V$. We have
$$h_1(0)=0<h_1(x_0)=f(x_0) > h_1(y_0)=f(y_0)-2[f(y_0)-f(x_0)]=f(x_0)-[f(y_0)-f(x_0)].$$
So the function $h_1$ is not monotone.
$\hspace{2cm}\square$
The basic idea of the proof of this lemma is that if we have function which look similarly to the functions in the following picture, we can find a linear combination, which is not monotone.


Corollary.
If $\Zobr{f,g}{\intrv01}{\R}$ are functions such that the function $af+bg$ is monotone for any $a,b\in\R$, then $f(x)=c$ for some constant $c\in\R$
or $g(x)=cf(x)+d$ for some constants $c,d\in\R$.
Proof.
We apply the above lemma to the functions $f_1(x)=f(x)-f(0)$ and $g_1(x)=g(x)-g(0)$. $\hspace{2cm}\square$

The claim of the exercise follows from this corollary. Indeed, suppose that $V$ is a subspace of $\R^{\intrv01}$ which contains only monotone functions. We can assume that $V$ contains all constant functions, since adding a constant function does not influence monotonicity. The corollary says that if we take two linearly independent functions $1,f\in V$, then all remanding functions in $V$ are linear combinations of $1$ and $f$.

Remark. The above proof can be adapted without much effort to
functions from $\mathbb R$ to $\mathbb R$ (instead of $[0,1]\to\mathbb
R$). We just need to add one more case $y_0<0<x_0$ to our lemma. (It
is sufficient to deal only with $x_0>0$, since the case $x_0<0$ is
symmetric.)
A: If $V$ is a vector space of monotone functions and $f_1, f_2, f_3 \in V$ then $\left(f_i(0), f_i(1)\right) \in \mathbb{R^2}$ are three vectors in a two-dimensional space and therefore dependent. That means that there is a non-trivial linear combination of $f_1, f_2, f_3$ that vanishes at $0$ and $1$ and because it is also monotone, it is identically zero. So any three elements of $V$ are linearly dependent.
A: Somewhat simpler: Let $V$ be a $3$-dimensional vector space of monotone functions.  Consider any linearly independent $f_1,f_2 \in V$.  Then $g = (f_2(1) - f_2(0)) f_1 - (f_1(1) - f_1(0)) f_2 \in V$ has 
$g(0) = g(1)$, so since $g$ is monotone it must be constant.
Moreover since $f_1$ and $f_2$ are linearly independent, $g \ne 0$.  Thus the constant function $1 \in V$.  But if $V$ is $3$-dimensional, it has a basis that contains $1$, and we just showed this is impossible because $1$ is in the span of the other two basis elements.
EDIT: another way to say this. Suppose $f_1$, $f_2$, $f_3$ are linearly independent members of $V$.  We may assume $f_2$ is not a scalar multiple of $1$.  Then $1 = a f_1 + b f_2 = c f_2 + d f_3$ for some scalars $a,b,c,d$, with $a \ne 0$.  So $a f_1 + (b-c) f_2 - d f_3 = 0$, contradicting  linear independence. 
