Vector Space of Continuous on Closed Interval Real Functions is not Finite Dimensional

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Let $I := \closedint 0 1$ be a closed real interval.

Let $\struct {\map C I, +, \, \cdot \,}_\R$ be the continuous on closed interval real function vector space.

Then $\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional.


Monomials are linearly independent

Let $d \in \N_{>0}$.

Consider the set of real monomials of the following form:

$\map {x_n} t = t^n$

where $n \in \N_{>0}$ and $n \le d$.

Aiming for a contradiction, suppose the set of $x_n$ is not linearly independent.


$\forall n \in \N_{>0}: n \le d: \exists \alpha_n \in \R: \neg \forall n: \alpha_n \ne 0$


$\ds \sum_{k \mathop = 1}^d \alpha_k t^k = 0$

Let $m \in \N_{>0}: m \le d$ be the smallest index such that $\alpha_m \ne 0$.


$\ds \forall t \in \closedint 0 1: \sum_{k \mathop = m}^d \alpha_k t^k = 0$


$\ds \forall t \in \hointl 0 1: \sum_{k \mathop = m}^d \alpha_k t^{k - d} = 0$

Note that:

$\ds \forall n \in \N_{>0}: \exists t \in \hointl 0 1: t = \frac 1 n$


$\ds \forall n \in \N_{>0}: \sum_{k \mathop = m}^d \frac {\alpha_k} {n^{d - k} } = 0$

Passing the limit $n \to \infty$ gives us $\alpha_d = 0$.

This is a contradiction.

Hence, the set of $x_n$ is linearly independent.


$\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional

Aiming for a contradiction, suppose $\struct {\map C I, +, \, \cdot \,}_\R$ is finite dimensional and has the dimension $d$.

Any independent set of cardinality $d$ in a $d$-dimensional vector space is a basis for this vector space.

Then the set of monomials $x_n$ with $n \in \N_{>0}$ and $n \le d$ is a basis for $\struct {\map C I, +, \, \cdot \,}_\R$.

The constant function $\map x t = 1$ belongs to $\map C I$.


$\ds \forall n \in \N_{>0}: n \le d: \exists \beta_n \in \R: 1 = \sum_{k \mathop = 1}^d \beta_k \map {x_k} t$

Let $t = 0$.

Then $1 = 0$.

This is a contradiction.

Hence, $\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional.