Let $U$ be bounded, with a $C^1$ boundary. Show that a ''typical'' function $u\in L^p(U) (1\leq p &lt; \infty)$   does not have a trace on $\partial U$. More precisely there does not exist a bounded linear operator   $$T:L^p(U)\rightarrow L^p(\partial U) $$   such that $Tu=u|_{\partial U}$   whenever    $u\in C(\bar{U})\cap L^p(U)$
This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted  previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows:     Let $U$ be bounded with a $C^1$ boundary. Show that a ''typical''   function $u \in L^p(U) \ (1 \leq p &lt; \infty)$ does not have a trace on   $\partial U$. More precisely, prove there does not exist a bounded   linear operator      \begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation}      such that $Tu = \left. u \right|_{\partial U}$ whenever $u \in C(\overline{U}) \cap L^p(U)$.   So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then $Tu = \left. u \right|_{\partial U}$ is undefined.  I like the idea of using the $\mathrm{dist}(x, \partial U)$ function to do this. I was thinking about defining the function  \begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation}  which must be in $L^p(U)$ since $U$ is bounded (hence integral of $\varepsilon$ over $U$ is finite, right?). But this function is not continuous at the boundary, and so $u \notin C(\overline{U})$. Can this be modified somehow?  Another option is to use that the boundary is $C^1$. So for each $x^0 \in \partial U$ there exists an $r &gt; 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n &gt; \gamma(x_1,\ldots, x_n) \}$. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the $C^1$ function $\gamma$. How can we use this?
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