Let $T,L> 0$ two real numbers and we consider the Sobolev space $X := L^2(0,T; H^1(0,L))\cap H^{1}(0,T;H^{-1}(0,L))$. My question is:

Given $f \in X$, the trace $ t \mapsto f(t,L)$ belongs to what space? Could someone indicate me a reference?


1 Answer 1


The spatial evaluation (or trace) operator $\mathrm{tr}_L$ at $L$ is well defined and continuous on $H^s(0,L)$ for $s>1/2$; a classic reference is [LM, Chapter 1.9]. (Of course the range $s>1/2$ is exactly the one for which $H^s(0,L)$ is embedded into $C([0,L])$.)

Thus, an easy immediate answer would be to apply $\mathrm{tr}_L$ to your function $f \in L^2(0,T;H^1(0,L))$ in the sense of $f(t,L) = \mathrm{tr}_L[f(t)]$ which would give $$\bigl[t \mapsto f(t,L)\bigr] \in L^2(0,T).$$

If you want to squeeze a bit more: It is a classical result that $X \hookrightarrow C([0,T];L^2(0,L))$, see e.g. [LM, Chapter 1.3.1]. In particular, $X \hookrightarrow L^p(0,T;L^2(0,L))$ for every $p \in [1,\infty)$; so, if $f \in X$, then $$f \in L^p(0,T;L^2(0,L)) \cap L^2(0,T;H^1(0,L)) \hookrightarrow \Bigl[L^p(0,T;L^2(0,L)),L^2(0,T;H^1(0,L))\Bigr]_s$$ for all $s\in (0,1)$ and $p \in [1,\infty)$, where the space $Y_s$ on the right is the complex interpolation space. Since the (Bochner) Lebesgue spaces are compatible with complex interpolation as in [BL, Theorem 5.1.2] and $H^s(0,L) = [L^2(0,L),H^1(0,L)]_s$ (see again [LM, Chapter 1.9]), the interpolation space $Y_s$ is $$Y_s= L^q\bigl(0,T;[L^2(0,L),H^1(0,L)]_s\bigr) = L^q(0,T;H^s(0,L)), \quad \frac1q = \frac{1-s}p + \frac{s}2.$$

As noted above, $\mathrm{tr}_L$ is well defined (and continuous) on $H^s(0,L)$ for $s > 1/2$, so by sending $p \to \infty$ and $s \searrow \frac12$, we obtain that $$\bigl[t \mapsto f(t,L)\bigr] = \bigl[t \mapsto \mathrm{tr}_L [f(t)]\bigr] \in L^{4-\varepsilon}(0,T)$$ for every $\varepsilon > 0$.

[BL] Bergh, Jöran; Löfström, Jörgen, Interpolation spaces. An introduction, Grundlehren der mathematischen Wissenschaften. 223. Berlin-Heidelberg-New York: Springer-Verlag. X, 207 p. with 5 figs. DM 60.00; $ 24.60 (1976). ZBL0344.46071.

[LM] Lions, J. L.; Magenes, E., Non-homogeneous boundary value problems and applications. Vol. I. Translated from the French by P. Kenneth, Die Grundlehren der mathematischen Wissenschaften. Band 181. Berlin-Heidelberg-New York: Springer-Verlag. XVI,357 p. DM 78.00 (1972). ZBL0223.35039.

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    $\begingroup$ My interpolation theory is a bit rusty and my copy of B&L is in my office: is it necessary to go through the intermediate interpolation for $X\hookrightarrow C([0,T]; L^2)$? Can one not directly take the complex interpolant of $[L^2_t H^1_x, H^1_t H^{-1}_x]_s$? $\endgroup$ Jan 25, 2021 at 14:47
  • $\begingroup$ You are right, it is not really necessary. I did try to rely on the "best known" results and not open the can of worms of vector-valued interpolation in the differentiability index (and embeddings). I thought one did not lose anything there since the $\varepsilon$ missing at the end would occur anyway, but I just realized that seemingly may not occur in the Hilbert space case according to the Lions/Magenes book. I will check further. $\endgroup$
    – Hannes
    Jan 25, 2021 at 15:13
  • $\begingroup$ Please do! I don't have a copy of Lions--Magenes and would love to know if there's a strengthening. (But yes, I entirely agree that vector valued interpolation with differentiability index is a big can of worms. Didn't B&L deliberately omit all but the simplest cases?) $\endgroup$ Jan 25, 2021 at 15:19
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    $\begingroup$ I guess all the vector valued stuff is not enough as long as one needs $H^s(0,L)$ with $s>1/2$ for the trace; even if we could interpolate the vector-valued function spaces in an optimal way, the obstruction $s>1/2$ will still produce the $4-\varepsilon$. $\endgroup$
    – Hannes
    Jan 25, 2021 at 15:36

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