# Dimension of Continuous functions

I think that $\{\text{continuous functions on } [0,1]\}$ has a different dimension from $\{\text{continuous functions on } [0,1]:F(0)=0, F\in C^1(0,1)\}$ due to the constraints in the latter. But how do I prove this?

(I realize that this question may be similar to the question Bases of spaces of continuous functions I asked previously, but this is hopefully more concise and I have thought about @Florian's enlightenment that the dimension of the set of continuous functions on a closed interval is infinite.)

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@Davide: Unfortunately, I am not familiar with Hamel basis. I am just trying to show that those 2 vector spaces are not isomorphic. (Based on a hunch ;-) ). –  jon b Sep 11 '11 at 19:59

Your intuition about the constraints may be leading you astray here. For example, if you took the continuous functions (an infinite dimensional space) and just added a constraint like $F(0)=0$, intuitively you would think the dimension of the new space would be "one less". But what's one less than infinity?
Let $E$ the vector space of the continuous functions on $\left[0,1\right]$ with the norm $\displaystyle\lVert f\rVert:=\sup_{0\leq x\leq 1}|f(x)|$ and $F$ the space of the continuous functions on $\left[0,1\right]$, which vanish at $0$ and which are continuously differentiable on $(0,1)$ (and with continuous derivative on $\left[0,1\right]$) with the norm $\displaystyle\lVert g\rVert:=\sup_{0<x<1}|g'(x)|$ (I put norms since the question is tagged functional-analysis, and in order to show that we can choose the isomorphism I gave on comment continuous for these (natural) norms). Let $\varphi \colon E\to F$ defined by $\displaystyle\varphi(f)(x)=\int_0^xf(t)dt$. $\varphi$ is linear, $\varphi(f)\in F$ for all $f\in E$, if $\varphi(f)=0$ then $\varphi(f)'=0$ hence $f$ is constant and is $0$, and $\varphi$ is onto since if $g\in F$ then $\varphi(g')=g$. Since $\displaystyle\lVert\varphi(f)\rVert_F=\sup_{0<x<1}\left|f(x)\right|\leq \lVert f\rVert_E$, $\varphi$ is continuous.