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The following fact seems to be well-known and easy to prove: Let $\sigma$ be a Borel measure on a Borel set $\Omega \subseteq \mathbb{R}^d$, then $$\|\sigma \|_{\smash{\dot{H}}^{-1}}:\,=\sup\limits_{\|f\|_{\smash{\dot{H}}^1}\leq 1} |\langle\sigma ,f\rangle| < \infty \, \Rightarrow \sigma (\Omega )=0 \, ,$$ where $\|f\|_{\smash{\dot{H}}^1}:\,=\|\nabla f\|_2$, and the inner product is just the integral of the product.

However, I couldn't find a proper reference in standard Sobolev Spaces textbooks (I tried Adams & Fournier and Leoni). Is this wrong? Does anyone know a solid reference?

(cross posted from MSE after a week with no answer).

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    $\begingroup$ This seems completely trivial: since $\|\cdot\|_{\dot{H}^1}$ is only a seminorm and $\|1\|_{\dot{H}^1} = 0$, then it follows immediately from the definition that in order to have $\|\sigma\|_{\dot{H}^{-1}} < \infty$ we must have $\langle \sigma, 1 \rangle = 0$. If it's that simple, I wouldn't think any reference should be needed. $\endgroup$
    – Nate Eldredge
    Commented Aug 7, 2019 at 19:58
  • $\begingroup$ I agree it's not too hard, but a reference would be helpful for my readers. $\endgroup$
    – Amir Sagiv
    Commented Aug 8, 2019 at 13:44
  • $\begingroup$ @AmirSagiv, you might have not noticed, but your profile picture does not get displayed correctly. $\endgroup$
    – Alex M.
    Commented Sep 9, 2019 at 18:52
  • $\begingroup$ Thanks @AlexM. But It looks right to me. What's the issue? $\endgroup$
    – Amir Sagiv
    Commented Sep 9, 2019 at 18:54
  • $\begingroup$ @AmirSagiv: Instead of your profile image, I see a "broken file icon" image, identical to the one displayed in this question on SO. If, on the other hand, I right click on it and choose "View image", then your correct image gets displayed. I don't understand. I shall probably ask about it on Meta. $\endgroup$
    – Alex M.
    Commented Sep 10, 2019 at 18:45

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