A compact operator is associated with 'smoothing', i.e. a function will have more regularity after a compact operator has acted upon it.
So if we take $\phi \in H^1([0,1])$ and let $I:H^1([0,1]) \to L^2([0,1])$ act on it, where $I$ is the compact embedding operator, in what way does $I\phi = \phi$ have more regularity? Because the function still 'looks' the same? All we seem to have done is change the norm from the $H^1$ norm to the $L^2$ norm..so in what way has the function been smoothed or regularized by this embedding operation?