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In (function space) interpolation theory, a Banach space $E$ is of class $J(\theta)$ (for $0 < \theta < 1$) if $$X \cap Y \subseteq E \subseteq X+Y,$$ where $(X,Y)$ are Banach spaces and form an interpolation couple, and there exists a constant $C>0$ such that

$$\|x\|_E \leq C \|x\|_X^{1-\theta} \|x\|_Y^\theta \quad \text{for all}~x \in X\cap Y.$$

It is known that if $E$ is of class $J(\theta)$, then $(X,Y)_{\theta,1} \hookrightarrow E$, and every real or complex interpolation space of parameter $\theta$ is of class $J(\theta)$.


Now I am interested in a sort of reversion of this property in the case of Besov spaces $B^\theta_{p,1}(\Omega)$, where $X = L^p(\Omega)$ and $Y = W^{1,p}(\Omega)$ for $1 < p <\infty$ and $\Omega$ being either $\mathbb{R}^d$ or a bounded extension domain in the $L^p$- and $W^{1,p}$-sense. More specifically:

Question: Let $(f_n) \subset W^{1,p}(\Omega)$ be a sequence which converges to zero in $$(L^p(\Omega),W^{1,p}(\Omega))_{\theta,1} = B^\theta_{p,1}(\Omega).$$ Does this imply that $$\|f_n\|_{L^p(\Omega)}^{1-\theta}\|f_n\|_{W^{1,p}(\Omega)}^\theta \to 0 \quad \text{as}~n\to\infty~\text{?}$$


The motivation to study this question is from Corollary 2 in Chapter 1.4.7 in Maz'ja's Sobolev spaces where it is established that $$\|\operatorname{tr} f\|_{L^q(\partial\Omega)} \leq C \|f\|_{L^p(\Omega)}^{1-\theta}\|f\|_{W^{1,p}(\Omega)}^\theta \quad \text{for all}~f \in W^{1,p}(\Omega)$$ for suitable combinations of $p,q$ and $\theta$; $\operatorname{tr}$ being the trace operator.

I would like to use this inequality to extend the trace operator from $W^{1,p}(\Omega)$ to $B^\theta_{p,1}(\Omega)$ for which I need to show that the operator is closable there, i.e., if $(f_n) \subset W^{1,p}(\Omega)$ converges to zero in $B^\theta_{p,1}(\Omega)$ and $(\operatorname{tr} f_n)$ converges to some $g \in L^q(\partial \Omega)$, then already $g = 0$.

Due to the real and complex interpolation spaces of parameter $\theta$ between $L^p(\Omega)$ and $W^{1,p}(\Omega)$ being of class $J(\theta)$, the property to be shown would also imply that $(f_n) \to 0$ in such spaces; this however is in compliance with the property that $B^\theta_{p,1}(\Omega)$ already continuously embeds into all those.


I have tried, amongst other attempts, using the characterization of Besov spaces via an integrated weighted modulus of smoothness (see e.g. this related question) to show that difference quotients of $f_n$ and thus $\|\nabla f_n\|_{L^p(\Omega)}$ stays bounded, but this did not work and I think this claim is too strong (the argument would imply that every convergent sequence in the interpolation space on $\Omega$ also converges weakly in $W^{1,p}(\Omega)$, if I am not mistaken). In the expression to be shown to converge to zero, $\|f_n\|_{W^{1,p}(\Omega)}^\theta$ is allowed to be unbounded in general, because $(f_n)$ goes to 0 in $L^p(\Omega)$ in any case.

Please also note that the way I arrived at the question at hand might be sub-optimal in the sense that I know that the closedness-property of $\operatorname{tr}$ is sufficient for what I want to ultimately show, but might not be the weakest condition (it is only the weakest I could come up with..). It might thus be well possible that the question has a negative answer.

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Here is a negative answer for a certain range of $p$ and $\theta$. It shows that one can have convergence to zero in a $\gamma$-Besov norm, where $\gamma$ may be larger than $\theta$. Hope I didn't mess up the parameters.

Claim. For any $p\in (1,\infty)$ and $\theta, \gamma \in (0, 1 - 1/p)$ there are $f_\varepsilon\in W^{1,p}(0,1)$ such that, as $\varepsilon \to 0$, $$\|f_\varepsilon\|_{B_{p, 1}^\gamma(0,1)} \to 0, \qquad \|f_\varepsilon\|_{L^p(0,1)}^{1-\theta} \|f_\varepsilon\|_{W^{1, p}(0,1)}^\theta \to \infty.$$

Proof of the claim. Fix $p$ and $\theta$ as above, and let $\varepsilon > 0$ be small. For $\alpha \in (\theta, 1 - 1/p)$ and small $\beta >0$ define $$f_\varepsilon(x) = \varepsilon^\beta (x+\varepsilon)^\alpha, \qquad x\in (0, 1).$$ Since $f_\varepsilon \in C^1[0,1]$, we have $f_\varepsilon \in W^{1,p}(0,1)$ for all $\varepsilon$. We calculate $$\|f_\varepsilon\|_{L^p(0,1)}^p = \varepsilon^{p\beta} \int_0^1(x+\varepsilon)^{p\alpha} dx = \frac{\varepsilon^{p\beta}}{p\alpha + 1} \left ((1+\varepsilon)^{p\alpha + 1} - \varepsilon^{p\alpha + 1}\right),$$ which implies $\|f_\varepsilon\|_{L^p(0,1)} \sim \varepsilon^\beta.$ Moreover, $$\|f_\varepsilon'\|_{L^p(0,1)}^p = \varepsilon^{p\beta} \alpha^p \int_0^1(x+\varepsilon)^{p(\alpha -1)} dx = \frac{ \varepsilon^{p\beta} \alpha^p}{p(\alpha -1) + 1} \left ( (1+\varepsilon)^{p(\alpha -1) + 1} - \varepsilon^{p(\alpha -1) + 1} \right ).$$ The choice $\alpha < 1 - 1/p$ yields $\|f_\varepsilon\|_{W^{1, p}(0,1)} \sim \varepsilon^{\beta + \alpha -1 + 1/p},$ whence $$\|f_\varepsilon\|_{L^p(0,1)}^{1-\theta} \|f_\varepsilon\|_{W^{1, p}(0,1)}^\theta \sim \varepsilon^{(1-\theta)\beta + \theta(\beta + \alpha -1 + 1/p)} = \varepsilon^{\beta + \theta(\alpha - 1 + 1/p)}.$$ This is unbounded for $\varepsilon \to 0$ if $\beta$ is sufficiently small, as $\theta(\alpha - 1 + 1/p) <0$ by the choice of $\alpha$.

On the other hand, for $\eta \in (0, \alpha)$ we have $$\|f_\varepsilon\|_{C^\eta[0,1]} = \|f_\varepsilon\|_\infty + \varepsilon^\beta \sup_{x, x+h\in [0, 1]} \frac{|(x+ h + \varepsilon)^\alpha - (x+\varepsilon)^\alpha|}{|h|^\eta} \sim \varepsilon^\beta,$$ as $t \mapsto t^\alpha$ is $\eta$-Hölder continuous on $[0, 1]$ if $\eta < \alpha$. Since $C^\eta \hookrightarrow B_{p,1}^{\gamma}$ for $\gamma \in (0, \eta)$, we conclude that $$\|f_\varepsilon\|_{B_{p, 1}^\gamma(0,1)} \leq C \|f_\varepsilon\|_{C^\eta[0,1]} \sim \varepsilon^\beta \to 0.$$ Because we can choose $\alpha$ as close to $1-1/p$ as we want, the same is true for $\eta$ and thus for $\gamma$ as well.

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  • $\begingroup$ Thanks! A late but welcome surprise of an answer. I need to ponder about this for a little bit. Maybe you could give me a hint where the relation between $\theta$ and $\alpha$ is used? Right now it seems to me like the argument should work for any $\theta>0$ up to (including) $1$. $\endgroup$
    – Hannes
    Aug 24 '20 at 13:36
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    $\begingroup$ Indeed, the upper bound on $\theta$ is nowhere used in the proof of the claim. However, $\gamma < 1 - 1/p$ is used for convergence to zero in a Hölder/Besov norm. So only for $\theta < 1-1/p$ the claim gives a counterexample to your question (if the calculations are correct). I introduced $\gamma$ to show that the failure of the implication in OP is not "sharp" w.r.t. $\theta$. For example, even if $\theta$ is small and $\gamma$ is close to $1-1/p$, there is a choice of $\alpha, \beta$ which yields to a counterexample. $\endgroup$
    – phantomias
    Aug 24 '20 at 14:05
  • $\begingroup$ That's what I figured. In hindsight, checking with scaled power/root functions seems kind of immediate now, as always. Thanks again. $\endgroup$
    – Hannes
    Aug 24 '20 at 14:15
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This is a well known and classical argument of Lions & Peetre and has been used by numerous mathematicians since. An up to date proof can also be found here on page 124/125.

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    $\begingroup$ Since the question gets bumped to the front page every now and then, I want to add that, yes, the corresponding general statement about interpolation spaces is claimed in the book of Luc Tartar as mentioned in this answer. I however think that there is a gap in the proof. I have written to Luc Tartar already but we have not been able to clarify the problematic step in the proof so far. (And personally I doubt that it will be true without additional assumptions on the operator such as the one in my question..) $\endgroup$
    – Hannes
    Nov 25 '18 at 13:03

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