$f=0$ in $H^{-1}(\Omega)$ implies $f=0$ almost everywhere

Question

Does $f=0$ in $H^{-1}(\Omega)=(H^1_0(\Omega))^*$ implies $f=0$ almost everywhere in $\Omega$?

Answer 1

A bit of a pet peeve of mine: the negative Sobolev spaces are spaces of distributions. However, your question (phrased as asking $f = 0$ a.e.) presupposes that elements of $H^{-1}$ can be represented as functions. But that's not always the case.

Example: Let $\Omega = (0,1)\subsetneq \mathbb{R}$. Then $H^1_0(\Omega) \hookrightarrow C^{1/2}(\Omega)$, and hence the Dirac $\delta$ distribution acts continuously on $H^1_0(\Omega)$. So $\delta \in H^{-1}(\Omega)$.

It is not very meaningful to ask whether a distribution (say $\delta$) is "almost everywhere zero.

Since the elements of $H^{-1}$ are automatically distributions, you have that $f= 0$ in $H^{-1}$ implies that $f$ is the zero distribution.

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Now, arbitrary functions do not represent distributions. So a reasonable interpretation of your question is:

If $f\in L^1_{\mathrm{loc}}(\Omega)\cap H^{-1}(\Omega)$ is 0 in $H^{-1}(\Omega)$, then does $f$ vanish a.e.?

In view of the discussion above the break, this reduces to asking whether a representation of the zero distribution by locally integrable functions is unique (up to equality a.e.), and the answer is easily seen to be yes.