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Would someone be able to tell me what the dimension is of the following vector space?

The space consists of all $C^2$ functions $f: [a,b]\longrightarrow\mathbb{C}$ which, for pre-specified real $\alpha_1,\alpha_2,\beta_1,\beta_2$, meet the conditions:

$$\alpha_1f(a)-\alpha_2f'(a)=0,\qquad \beta_1f(b)-\beta_2f'(b)=0$$

I hope you can help. Thank you.

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    Are you missing the actual differential equation that is satisfied by the functions? As stated it is infinite dimensional.2011-04-23
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    @Willie The space I've described is the range of a resolvent operator which is constructed to be the inverse of the operator $T-\lambda$ where where $T$ is a differential operator (with the above space its range). My aim is to show that $(T-\lambda)^{-1}$ is an operator of infinite rank, in order to get an infinite sequence of eigenvalues converging to zero (rather than a finite one). As you can see, the context was well worth omitting in the interests of getting an answer!2011-04-23

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Your space is the kernel of the linear map $C^2 \to \mathbb{C}^2$ given by $f \mapsto \begin{bmatrix} \alpha_1 f(a) - \alpha_2 f'(a) \\\ \beta_1 f(b) - \beta_2 f'(b)\end{bmatrix}.$ Since $C^2$ is obviously infinite-dimensional (the dimension is $\mathfrak{c} = 2^{\aleph_0}$), the kernel of this map has co-dimension at most two, so it must be infinite-dimensional as well.

As Willie states in his comment, this question becomes much more interesting when the functions $f$ are subject to some differential equation.

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    @Theo: Correct me if I'm wrong, but $C^2$ is a subspace of all continuous functions from $[a,b]$ to $\mathbb{C}$. Isn't that space of countable dimension (namely it is determined by values at rational points)2011-04-23
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    @Asaf: Yes, a function is determined by its values at the rational points, but that doesn't mean that the dimension is countable (this doesn't yield a basis). It suffices to exhibit a one-parameter family of linearly independent functions and that isn't too hard.2011-04-23
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    @Asaf: see also Jonas Meyer's nice answer.2011-04-23
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    @Theo: Ah, you are correct sir. It is funny how dealing with cardinalities within set theory is fine, but once you go to analysis I tend to get confused. :-) Thanks!2011-04-23
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    @Asaf: For me it's rather the other way around :)2011-04-23
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    @Asaf: For example, $\{e^{ax}:a\in\mathbb{C}\}$ is linearly independent.2011-04-23
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    @Jonas: I believe you meant $\{e^{tx}\mid t\in[a,b]\}$, but the principle's the same. Thanks :-)2011-04-23
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    @Asaf: You're welcome. I meant (and still mean) what I wrote, so please explain if I've made a mistake. The set I named is linearly independent on any infinite set. (I just realized that I probably shouldn't have used "$a$" because it has another meaning in the problem. Sorry if that caused confusion.)2011-04-23
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    @Jonas: I misread you, that is all. I was thinking that $x\in\mathbb{C}$ (i.e. the domain is $\mathbb{C}$) while it can also be restricted to $[a,b]$ where it fits the problem. Anyway, I believe that what I wrote also counts as a linearly independent set whose cardinality is continuum. :-)2011-04-23
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    Do you mean "at most" two because the image of the map may be less than two-dimensional?2011-04-23
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    @Josef: Yes, exactly.2011-04-23
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    @Theo Sorry if I'm dense, but how could it be less than two-dimensional?2011-04-23
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    @Josef: If $\alpha_1=\alpha_2=0$ or $\beta_1=\beta_2=0$.2011-04-23
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    @Josef: Nothing excludes $\alpha_1 = \alpha_2 = 0$ (which yields co-dimension at most one). If in addition $\beta_1 = \beta_2 = 0$ then the map is simply zero.2011-04-23
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    Ah, yes I see :) Sorry, this may have been a rather elementary question but it was really useful to get a reply and the map you made from functions to complex numbers was very nice. Thanks a lot to all.2011-04-23
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    @Josef: Don't worry, this happens to all of us.. You're welcome, of course.2011-04-23
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As Theo pointed out with a nice linear algebra argument, it is infinite dimensional. This answer is just to point out some explicit infinite dimensional subspaces. For example, if $[a,b]=[-1,1]$, your space contains all functions of the form $f(x)=e^{-1/(1-x^2)}p(x)$ where $p$ is a polynomial function (and $f(-1)=f(1)=0$). The set $\{e^{-1/(1-x^2)},e^{-1/(1-x^2)}x,e^{-1/(1-x^2)}x^2,e^{-1/(1-x^2)}x^3,\ldots\}$ is linearly independent. The case of arbitrary $[a,b]$ follows by a linear change of variables.

If $abump functions.

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    Thanks very much! That's a really nice visualisation of it.2011-04-23
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    @Josef: You're welcome. I realized after posting that taking $a$[a,b]$ to $[a,b]$ would also do, the point being that if $f$ is such a function then $f(a)=f'(a)=f(b)=f'(b)=0$. – 2011-04-23