# A typical $L^p$ function does not have a well-defined trace on the boundary

This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted here 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 $u \in L^p(U) \ (1 \leq p < \infty)$ does not have a trace on $\partial U$. More precisely, prove there does not exist a bounded linear operator

$$T:L^p(U) \to L^p(\partial U)$$

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

$$u(x) = \frac{1}{\mathrm{dist}(x,\partial U)},$$

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 $r > 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n > \gamma(x_1,\ldots, x_n) \}$. 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?

The key is that you can not find a bounded operator. Such an Operator is continuous. So what happen when you plug in a sequence of continuous functions?

Assume $\Omega=(0,1)$ and choose $$f_n (x):=\begin{cases} 1 &\text{ on } [0,1-1/n]\\ n-nx &\text{ on } [1-1/n,1]\end{cases}$$ So $f_n$ falls linearly to zero on $[1-1/n,1]$.

Each $f_n$ is a continuous function and has trace zero in $1$. Now we restrict the trace operator to $x=1$. Now $0=\lim_n T(f_n)\neq T(f)=1$ since $f_n\to 1$ in $L^2$.

• Brilliant! So since our bounded operator is continuous (we have iff between these two notions), then a sequence of continuous functions should yield its values at the boundary when acted on by the trace operator $T$. But this construction of functions yields a limit value of $T(f_n)$ that is different from $T(f)$, which is in contradiction with the fact that $T$ is continuous, right? – Numbersandsoon Mar 17 '13 at 9:41
• Exactly. And also not that $f_n$ is not uniformly bounded in $H^1$. In that case, due to the control of the norm of the derivatives, one has a continuous trace operator. – Quickbeam2k1 Mar 17 '13 at 10:16
• Ahh, very nice! Thank you for your help! I learned a lot from just this problem and your explanations. – Numbersandsoon Mar 18 '13 at 4:33
• Maybe what is also worth mentioning is that the example above does not work in $p=\infty$ since one does not have the uniform convergence of the sequence. Another point that you should note is that an $L^p$-function is precisely an element of an equivalence class of functions. Which means you can change the values of the $f_n$ on a set of measure zero without changing the trace of each of the $f_n$ (each $f_n$ is in $H^1$) – Quickbeam2k1 Mar 18 '13 at 9:32
• @Quickbeam2k1 What you presented in your answer seems to be a counterexample. But doesn't the problem ask to prove the nonexistence of a bounded linear operator for all $\Omega$ with a $C^1$ boundary, not just with $\Omega=(0,1)$? Can we instead assume that there does exist a bounded linear operator $T : L^p(U) \to L^p(\partial U)$ with $Tu=u|_{\partial U}$, and find a contradiction? – Cookie Jan 11 '15 at 19:46

For completeness, I'll expand the idea in this comment. The construction does not require $C^1$ boundary, and works in every bounded domain. Let $$u_n(x) = (1-n\operatorname{dist}(x,\partial U))^+$$ which is a continuous function on $\overline{U}$. Since the sequence $u_n^p$ is decreasing, it is dominated by $u_1^p$, which is integrable. Hence $$\lim_{n\to\infty}\int_U u_n^p = \int_U \lim_{n\to\infty} u_n^p = 0$$ On the other hand, $$\int_{\partial U}u_n^p = \int_{\partial U}1\not\to0$$ which yields the claim.

• Is $u_n^p$ simply the $u_n(x)$ you defined and raised to the $p$ power? If so, then I am trying to see why $u_n(x)=[(1-n\operatorname{dist}(x,\partial U))^+]^p$ is decreasing. I think after that, you recognized that $|u_n|^p \le u_1^p$, one of the conditions that allow you to use the Dominated Convergence Theorem to assert $\lim_{n \to \infty} \int_U u_n^p = 0$. – Cookie Jan 14 '15 at 3:25
• Decreasing with respect to $n$, meaning $u_{n+1}^p\le u_n^p$. Because each $u_n$ takes values between $0$ and $1$, it suffices to check $u_{n+1}\le u_n$. Which is not hard: in $u_{n+1}$ we subtract more. – user147263 Jan 14 '15 at 3:28
• Also, in your example, is it necessary to put in the "$+$" in $(1-n \operatorname{dist}(x,\partial U))^+$? Is that so $u_n(x)$ is postive for each $n$? I'm assuming that if I didn't put the $+$, then that function would no longer work. – Cookie Jan 14 '15 at 3:33
• Yes, this is the main point. The function should be zero when the distance to boundary is more than $1/n$. Without $+$ it would be something negative. – user147263 Jan 14 '15 at 3:40
• And we shouldn't use absolute values either because if $\operatorname{dist}(x,\partial U) > \frac 1n$, the function $u_n$ would be postive due to the absolute value and not $0$. If we just use the "$+$" like you introduced, then the negative part of the function $u_n$ would just simply be $0$. I hope that is correct. – Cookie Jan 14 '15 at 4:14

I like your idea of taking a function $\gamma:\mathbb R^+_0\to\mathbb R_0^+$ and composing with the function $d=\operatorname{dist}(\cdot,\partial U)$. This is how your idea will actually work:

You picked $\gamma(t)=1/t$ but that may be too singular to guarantee $L^p$, however $\gamma(t)=[\log 1/t]_+$ should work (possibly using the coarea formula).

To cure the fact that $\gamma(d(\cdot))$ has no trace, you may cut it off by defining the sequence $u_n(x)=n\wedge\gamma(d(x))$, whose trace is constant $n$ and by Lebesgue dominated convergence (or Fatou-Lebesgue theorem) all integrable with integral bounded by that of $\gamma(d(\cdot))$.