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In a paper, I've read the following thing. Here $\Omega$ is a smooth domain

From the standard trace theorem we know there exists a bounded linear operator $$\gamma: H^1(\Omega) \rightarrow H^{\frac{1}{2}}(\partial \Omega)$$ such that $\gamma_0 u=u_{|\partial W}$ for $u$ smooth and that satisfies $$\|\gamma_0 u\|_{H^{\frac{1}{2}}} \geq C \|u\|_{H^1} \qquad \forall u\in H^1(\Omega) \setminus \ker(\gamma_0)$$

I don't know how to show the last equality.

If I can prove that $\gamma_0$ has closed range, then I have that it's bounded from below ( by a standard result coming from Open Mapping Theorem). However, I don't know how to prove it. Any hint is highly appreciated!

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    $\begingroup$ I have doubts. Take a $u\in H^1_0(\Omega)$ and consider the sequence $u_\epsilon=u+\epsilon$ $\endgroup$
    – Piero D'Ancona
    Commented Oct 26, 2021 at 16:32
  • $\begingroup$ Piero's example is convincing, of course, but the inequality also doesn't make sense on general grounds, because we can always make the RHS large by giving $u$ extra oscillations in the interior of our domain. $\endgroup$
    – Christian Remling
    Commented Oct 26, 2021 at 17:23
  • $\begingroup$ You may want to assume $u\in (\ker \gamma_0)^\perp$. In this case the inequality follows from the open mapping theorem since $\gamma_0: (\ker \gamma_0)^\perp\to H^{1/2}$ is onto. Here $\perp$ refers to the orthogonal complement in the Hilbert space $H^1(\Omega)$ $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 26, 2021 at 17:40
  • $\begingroup$ @LiviuNicolaescu Thanks, I need a last clarification. How do you use the open mapping to show that bound? I really cannot see how $\endgroup$
    – bobinthebox
    Commented Oct 26, 2021 at 18:29
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    $\begingroup$ The map from $(\ker \gamma_0)^\perp\to H^{1/2}$ is bijective and continuous $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 26, 2021 at 21:58

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