$V$ a vector space of functions: $[0,1] \longrightarrow\mathbb{R}$ . let $U \subseteq V$, as U is a subset consists of mono' functions

by Jneven   Last Updated June 21, 2018 08:20 AM

Given the vector space $V$ of functions from $[0,1] \longrightarrow\mathbb{R}$, let $U$ be the subspace of $V$ consisting of monotone functions (every function $\in$ $U$ is monotone but not every monotone function is an object of $U$). What is the maximum dimension of $U$? (by “monotone” I'm referring to weakly Monotonically : if $x>y\longrightarrow f(x)\geq f(y)$ or if $x>y\longrightarrow f(x)\leq f(y)$.

I was thinking to approach this question using elementary set theory which I learned to find the cardinality of $U$, but that wouldn't be very possible on the interval $[0,1]$.

I've never seen a Q. like this so I don't know what else can be done.

Answers 2

Hint: If $f$ and $g$ are monotonic functions and none of them is a multiple of the other one, then there are real numbers $\alpha$ and $\beta$ such that $\alpha f+\beta g$ is not monotonic.

José Carlos Santos
José Carlos Santos
June 21, 2018 07:32 AM

The cardinality of the monotone real functions defined on $[0,1]$ is equal to $\mathfrak{c}$. See What is the cardinality of a set of all monotonic functions on a segment [0,1]?

Hence the dimension of $U$ is at most $\mathfrak{c}$. Now you'll be able to prove that the family of functions

$$f_\alpha(x)= \begin{cases} 0 & 0 \le x <\alpha\\ 1 & \alpha \le x \le 1 \end{cases}$$ for $\alpha \in (0,1)$ is linearly independent. Hence, the dimension of $U$ (over $\mathbb R$) is equal to $\mathfrak{c}$ (while the dimension of $V$ is equal to $2^{\mathfrak c}$).

June 21, 2018 08:01 AM

Related Questions

Decreasing function depending in parameters

Updated January 22, 2018 06:20 AM

Is h an element of the linear space $F(Ω,R)$?

Updated January 31, 2018 22:20 PM

If $T\circ S=id$ then $ker(S)=\{0\},ker(T)=\{0\}$

Updated February 07, 2018 19:20 PM

Is this really a vector space?

Updated February 11, 2018 14:20 PM