TITLE: How to think intuitively about compact injections?
QUESTION [6 upvotes]: Motivation is the Sobolev inequalities, but there are more basic examples. For example, consider the inclusion 
$$i: C^1([0,1]) \rightarrow C([0,1])$$ 
This is compact since if $\{f_n\} \subset B_1(0,1)$, then for all $x, y \in [0,1]$ we have 
$$|f_n(x) - f_n(y)| \leq |f_n'|_{\infty}|x - y| \leq |x - y|$$
by the mean value theorem. So $\{i(f_n)\}$ is uniformly bounded and equicontinous, hence has a convergent subsequence. 
How should one think of a compact injection intuitively or informally?

REPLY [2 votes]: Depending what you mean by "intuitive", the Hilbert-space case of Sobolev spaces may have more to grab onto, and having spectral expansions makes it even more explicit. E.g., on the circle, $\sum_n c_ne^{inx}$ has $L^2$ norm-squared essentially $\sum_n |c_n|^2$, and has $H^1$ norm-squared $\sum_n (1+n^2)|c_n|^2$. Thus, an orthonormal basis for $H^1$ consists essentially of $e^{inx}/\sqrt{1+n^2}$. The images of these under the inclusion to $L^2$ are still orthogonal, but they're much smaller than the unit ball. In particular, the image of the unit ball from $H^1$ is a Hilbert-cube-like thing inside $L^2$, and is easily seen to be pre-compact (e.g., by the total-boundedness criterion).
Slight edit: and, for that matter, though it may seem less physically real, describing Sobolev spaces on the spectral side, on the circle, obviously converts things to Sobolev-like spaces inside $\ell^2$...