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Let $\Omega$ be a bounded Lipschitz domain.

Let $u \in H^1(\Omega) \cap L^\infty(\Omega)$, and suppose that $\lVert u \rVert_{L^\infty(\Omega)} \leq A$.

Let $T:H^1(\Omega) \to L^2(\partial\Omega)$ be the trace mapping.

Is it true that $Tu \in L^\infty(\partial\Omega)$ with $\lVert Tu \rVert_{L^\infty(\partial\Omega)} \leq A'$ for some constant $A'$?

I think so, since we can find functions $u_n \in C^0(\bar \Omega)$ bounded by $A$ such that $u_n \to u$ in $H^1$ and $Tu = \lim Tu_n$ in $L^2$.

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  • depending on how you define $T$, it can be obvious that it is also a bounded operator $L^\infty(\Omega) \to L^\infty(\partial \Omega)$ – reuns Feb 16 '16 at 15:00
  • @user1952009: Can you tell me how $T$ becomes a bounded operator from $L^\infty(\Omega)$ to $L^\infty(\partial\Omega)$? – gerw Feb 16 '16 at 20:32

1 Answers1

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Consider the function $u + A$. It belongs to $H^1(\Omega)$ and is non-negative. A standard procedure yields a sequence $\{v_n\} \in C(\bar\Omega) \cap H^1(\Omega)$ with $v_n \ge 0$ and $v_n \to u + A$ in $H^1(\Omega)$. Now, $T v_n \ge 0$, since it corresponds with the usual trace of $v_n$. Since $T$ is continuous, you have $T v_n \to T(u + A)$ and $T(u+A) \ge 0$. Now, you can easily show $T(u + A) = T u + A$ and this yields $T u \ge -A$. Similarly, $T u \le A$ follows.

gerw
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