Hahn-Banach theorem seems to permit "unpermitted" operations

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I've been taught to be very careful with the types of operations that we perform on functions in Sobolev spaces. Yet, the Hahn-Banach theorem seems to completely do away with the necessity of restraint. It says that any functional in the dualspace of a linear subspace may be extended to a functional in the dualspace of the complete space. If the subspace is dense, then this extension is even unique. Two examples that bother me:

  • The trace of the gradient is an unbounded operator in $H^1$, so the surface integral thereof is not a bounded linear functional. But on $H^2$ it would be, and this is a dense linear subspace. So Hahn-Banach gives us a unique extension of this operation to $H^1$.
  • The Dirac delta distribution is in the dualspace of $H^1_0([0,1])$, a dense linear subspace of $L^2([0,1])$. Per Hahn-Banach we are thus permitted to take point evaluations in $L^2([0,1])$.

These are two operations that I know are not allowed, so I must be misunderstanding the theorem or drawing the wrong conclusion. If we focus on the second example: are we really allowed to use Hahn-Banach to extend $\delta[\cdot]$ to $L^2([0,1])$? And if so, then what kind of extension would this produce if not a point evaluation?

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The misunderstanding here is that in these spaces your norms are changing. Remember that a set is dense in a topological space if the closure of that set in the topology equals the whole space, and this topology very much depends on the norm. So indeed the set $H^1_0$ is a dense subspace of $L^2$, when equipped with the $L^2$ norm (and the corresponding topology), but that does not mean that the usual $H_0^1$ equipped with the $H^1$ norm is a dense subspace of $L^2$ topologically speaking, because these are in fact two entirely different topological spaces, so that notion does not even make sense. The $H^1$ norm and $L^2$ norm are very different, so the dual of $H_0^1$ as a Banach space is different from the dual of the subspace of $L^2$ consisting of the vectors of $H_0^1$. These are the same vector spaces, but different normed spaces. The Hahn-Banach theorem allows us to extend linear functionals on the latter space to linear functionals on $L^2$, but not on the former.