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: 

> **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.