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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.)

Thanks loads!

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  • $\begingroup$ @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 ;-) ). $\endgroup$ – jon b Sep 11 '11 at 19:59
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As a general rule, once vector spaces are infinite dimensional, "dimension" becomes too crude a tool to tell them apart. Most infinite dimensional vector spaces that people use actually all have the same dimension. Therefore they are isomorphic, as vector spaces; but there is usually extra structure (such as a norm, or a topology) in which they differ.

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?

In fact, despite your intuition, your two vector spaces are isomorphic. The fact that one is a proper subset of the other doesn't prevent this. (Think of the fact that there's a one-to-one correspondence between the set of even integers and the set of all integers.) Knowing this, can you find an isomorphism?

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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.

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