Why is $C^5[a,b]$ infinite dimensional

Question

Let $V$ denote the vector space $C^5[a,b]$ over $R$.

How to show it is infinite dimensional?

I know that we can write:

$C^5[a,b]$ = { $f\in$ $C[a,b]$ : $5$th derivative exists and is continuous}

How to show that there does not exist a linearly independent subset of $V$ which spans V ?

Polynomials are in that vector space.. – Cameron Williams Jun 17 '20 at 20:24 — Cameron Williams, Jun 17 '20 at 20:24
https://math.stackexchange.com/a/1844631/269764 – Brevan Ellefsen Jun 17 '20 at 20:24 — Brevan Ellefsen, Jun 17 '20 at 20:24

score 8 · Accepted Answer · answered Jun 17 '20 at 20:25

8

Hint:

Note that $\mathbb{R}[x] \subset C^5([a,b],\mathbb{R})$ and that $\mathbb{R}[x]$ is not finite dimensional.

answered Jun 17 '20 at 20:25

markvs · Answer 2 · 2020-06-17T20:54:59.153

3

Clearly functions $x^n$, $n=1,2,...$ are in your space. To prove that these functions are linearly independent consider the Wronskian.

edited Jun 17 '20 at 20:54

answered Jun 17 '20 at 20:28

markvs

Yes..I can see they are linearly independent..How does that imply anything? – Gitika Jun 18 '20 at 10:42
1

That means that the space is inf. dim. A fin. dim. space cannot contain an infinite l.i. subset. – markvs Jun 18 '20 at 12:27