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 $J(\theta)$.




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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: 



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> **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 $$\|f\_n\|\_{L^p(\Omega)}^{1-\theta}\|f\_n\|\_{W^{1,p}(\Omega)}^\theta \to 0 \quad \text{as}~n\to\infty~\text{?}$$
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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.




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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](https://mathoverflow.net/questions/187527/whats-the-idea-behind-various-equivalent-norms-on-besov-spaces-bs-p-q)) 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.



"Reversion" of class $J(\theta)$ interpolation property for Besov spaces
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.



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> **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 $\varepsilon \to 0$,
> $$\|f\_\varepsilon\|\_{B\_{p, 1}^\gamma(0,1)} \to 0, \qquad \|f\_\varepsilon\|\_{L^p(0,1)}^{1-\theta} \|f\_\varepsilon\|\_{W^{1, p}(0,1)}^\theta \to \infty.$$
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**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.


