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.
 A: 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.
A: 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.
