Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
+1 for citing that book. :-) –  WimC Jun 8 '12 at 15:48
    
Maybe it's worth mentioning that if we solve this problem for monotone functions $[0,1]\to\mathbb R$, the same must be true for any closed interval. This can be used to answer the same question for monotone functions $\mathbb R\to\mathbb R$, since $\mathbb R=\bigcup_{n=1}^\infty [-n,n]$. –  Martin Sleziak Jun 11 '12 at 11:15
1  
Yes that's a nice extension although your last step ($\mathbb{R} = \cup \dotsc$) then still needs a little bit more explanation, since a single relation must work for all intervals. –  WimC Jun 11 '12 at 11:19
    
I was probably too hasty to write the above comment @WimC; Later I'll try to teg back to the question whether some argument along these lines can be done to get the result for $\mathbb R^{\mathbb R}$ instead of $\mathbb R^{[0,1]}$. –  Martin Sleziak Jun 11 '12 at 12:35

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer
1  
why is a function which is monotone and vanishes at 0 and 1 identically zero? –  Mariano Suárez-Alvarez Jun 8 '12 at 22:32
    
@Mariano The question was formulated for functions $[0,1]\to\mathbb R$. –  Martin Sleziak Jun 9 '12 at 3:29

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.

share|improve this answer
    
Nice! ${}{}{}{}{}{}$ –  leo Jun 9 '12 at 3:18

$\newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\R}{\mathbb R}\newcommand{\intrv}[2]{[{#1},{#2}]}$ 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.

f and h


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.)

share|improve this answer
    
can u tell me why is $h(x_0)=0$ in the first proof ? –  Theorem Jun 8 '12 at 14:51
    
i think $c$ is $ \frac {f(x_0)} {g(x_0)}$ –  Theorem Jun 8 '12 at 14:53
    
I've corrected the definition of $h(x)$; it should be $h(x)=g(x)-cf(x)$. Thanks for noticing this, @Ananda. –  Martin Sleziak Jun 8 '12 at 14:55
    
Can i post this problem on my blog ? Let me know if the norms of MSE allows to do so ? –  Theorem Jun 8 '12 at 15:03
    
@Ananda I don't see any problem with that. Moreover, the problem is taken from a book, it was not my own invention. Perhaps you can post link to your blog here, if you do so. –  Martin Sleziak Jun 8 '12 at 15:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.