Show that the real vector space of all continuous real valued functions on the interval $[0,1]$ is infinite dimensional.

attempt: Suppose there is a sequence of real valued functions on V, that is $f_1 , f_2 , f_3,....$ such that $f_1,f_2,....,f_m$ is linearly independent $\forall m\in \mathbb{N}$. So then there exists $a_1,...,a_m \in \mathbb{R}$ such that $a_1f_1 +a_2f_2 + ....+ a_mf_m = 0$, I am not sure how to continue . I am trying to show that all the coefficients are zero but I am not sure why and then conclude V is infinite dimensional . Can someone please help me? Thank you!

  • $\begingroup$ what course are you taking? $\endgroup$ – KonKan Sep 15 '16 at 4:39
  • $\begingroup$ advance linear algebra $\endgroup$ – user4294 Sep 15 '16 at 4:40

Notice that this vector space contains the space of all polynomials as a subspace, and $\{1, x, x^2, ... \}$ is linearly independent.


We'll show that $\{1, ..., x^n\}$ is linearly independent. Suppose that $p(x):= a_0 + ... + a_n x^n =0$ but all $a_i \neq 0$. Then $p(a) =0$ for all $a \in \Bbb R$, which implies that $p$ is the zero polynomial. This is absurd.

  • $\begingroup$ So $1,x,x^2,....,x^m$ is linearly independent and so all the coefficients must be zero. But I still don't see why that is. I think it has to do with the roots. $\endgroup$ – user4294 Sep 15 '16 at 4:35
  • $\begingroup$ @KonKan I think they're asking why it is the case that this set is linearly independent. The OP is right that it follows from a polynomial having only finitely many roots. $\endgroup$ – 3-in-441 Sep 15 '16 at 4:39

The linear map $f \mapsto x \cdot f(x)$ is injective: if a continuous function $f$ satisfies $xf(x) = 0$ for all $x \in [0,1]$, the you can conclude that $f(x) = 0$ for all $x$.

It is not surjective, because $x \cdot f(x)$ can only represent functions taking the value $0$ at $x=0.$

If your space were finite-dimensional then this would violate the rank-nullity theorem.


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