Does a set of 'm' linearly independent continuous functions constitute a Hilbert Space

Question

If I have a Sobolev space $\mathcal{H}^m[a,b]$ of functions $f : [a,b]\rightarrow\mathbb{R}$ where for all $f \in\mathcal{H}^m[a,b]$, $f$ and all derivatives up to order $m-1$ are absolutely continuous with the $m$'th derivatives of $f$ being square-integrable. If I denote a subspace of $\mathcal{H}_0^m[a,b] = \{f \in\mathcal{H}^m[a,b] \ | \ Lf = 0\}$ where $L$ is a linear, constant coefficient differential operator, then $\mathcal{H}_0^m[a,b]$ (the null-space of $L$) will consist of $m$ linearly independent functions $u_1, \ldots, u_m$. Given that $m < \infty$ we know that $\mathcal{H}_0^m[a,b]$ is of finite dimension, does this mean that $\mathcal{H}_0^m[a,b]$ as described is necessarily closed and therefore complete under some norm induced by an inner-product $\textit{i.e.}$ is it a Hilbert Space? I would also like to make the statement that all evaluation functionals on $\mathcal{H}_0^m[a,b]$ are continuous which would make it a reproducing kernel Hilbert space. I would like to make the overall statement that, given a linearly independent set of functions $u_1,\ldots,u_m \in \mathcal{H}$ that $\mathcal{H}$ is necessarily a reproducing kernel Hilbert space by virtue of the fact that it consists of a finite number of linearly independent functions. I am curious if I am stretching too far here with my statement. If not can anyone tell me what theorems or proofs I could used to make this statement in a paper I am writing more exact; otherwise, please let me know what I must add to make it correct.