Dimension of a vector space of functions subject to boundary conditions

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. –  Willie Wong Apr 23 '11 at 16:19
@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! –  Josef K. Apr 23 '11 at 16:26

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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@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. –  t.b. Apr 23 '11 at 16:24
@Asaf: For me it's rather the other way around :) –  t.b. Apr 23 '11 at 16:28
@Asaf: For example, $\{e^{ax}:a\in\mathbb{C}\}$ is linearly independent. –  Jonas Meyer Apr 23 '11 at 16:28
@Josef: If $\alpha_1=\alpha_2=0$ or $\beta_1=\beta_2=0$. –  Jonas Meyer Apr 23 '11 at 16:57
@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. –  t.b. Apr 23 '11 at 16:59
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 $a<c<d<b$, then your space also contains all smooth functions with support in $[c,d]$, which is infinite dimensional as a consequence of the existence of bump functions.
@Josef: You're welcome. I realized after posting that taking $a<c<d<b$ was being overly cautious. The restrictions of smooth functions with support in $[a,b]$ to $[a,b]$ would also do, the point being that if $f$ is such a function then $f(a)=f'(a)=f(b)=f'(b)=0$. –  Jonas Meyer Apr 23 '11 at 17:03