Let $f$ be a measurable function belonging to a Besov space of non-positive regularity $B^{\gamma}_{p,\infty}$, where $\gamma\le 0$, $p\in[1,\infty]$. Is it true that $|f|$ belongs to the same space and $\|\,|f|\,\|_{B^{\gamma}_{p,\infty}}\le C\|f\|_{B^{\gamma}_{p,\infty}}$ for some universal $C=C(\gamma,p)>0$?
Note that if $\gamma\in(0,1)$, then the answer is positive. Indeed, in this case $$ \|f\|_{B^{\gamma}_{p,\infty}}=\|f\|_{L_p(\mathbb{R}^d)}+\sup_{h\in[0,1]^d} h^{-\gamma}\|f(h+\cdot)-f\|_{L_p(\mathbb{R}^d)}, $$ and clearly $|\,|f(h+\cdot)|-|f|\,|\le |f(h+\cdot)-f\,|$. Thus, $\|\,|f|\,\|_{B^{\gamma}_{p,\infty}}\le \|f\|_{B^{\gamma}_{p,\infty}}$
I wonder whether the same is true for non-positive $\gamma$.
UPD: If $\gamma<0$, the answer is negative. The counter-example suggested by user Gerw is $f(x)=\sin nx$.
What if $\gamma=0$? Clearly, the answer is positive for the space $B^0_{2,2}=L_2$, but what about the space $B^0_{p,\infty}$?
I think that this is not possible for arbitrary elements. Let us consider a negative Besov space $B$. If $f \in B$ would imply $|f| \in B$, we could partition $f = f^+ - f^-$ with positive and negative part $f^{\pm} = (|f| \pm f)/2$. This implies $B = B^+ - B^+$ with $B^+ = \{ f \in B \;|\; f \ge 0\}$. However, elements in $B^+$ are measures, thus, $B^+ - B^+$ would consists only of signed measures. However, in most cases, $B$ should contain elements that are not signed measures, see, e.g., Decomposition of functionals on sobolev spaces for the case $B = H^{-1}(\Omega) = (H_0^1(\Omega))^\star$.
I am not so sure about the case of measurable functions $f$. But maybe one can modify the construction from Decomposition of measures acting on sobolev spaces to obtain a function $f \in H^{-1}(\Omega)$ with $f^+ \not\in H^{-1}(\Omega)$.