We know the standard trace inequality: for a bounded domain with certain boundary regularity, there is a $C>0$ such that

$$ \|Tu\|_{L^2(\partial \Omega)}\leq C\|u\|_{H^1(\Omega)}, \quad \quad u\in H^1(\Omega). $$

Here $Tu$ is the trace of $u$ on $\partial \Omega$. I roughly remembered that I encountered a more convenient version of this trace inequality with $\epsilon$ somewher. Namely, for any given $\epsilon>0$, it holds

$$ \|Tu\|_{L^2(\partial \Omega)}\leq \epsilon \|\nabla u\|_{L^2(\Omega)}+C(\epsilon)\|u\|_{L^2(\Omega)}, \quad \quad u\in H^1(\Omega). $$

Anyone can provide an elementary proof for this? Highly appreciated!


1 Answer 1


Take a look at the proof of the trace inquality, for example in Evans book


when he apply Young's inequality you could apply Young's inequality with epsilon.


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