Let $L>0$ fixed. Consider the space $$ \mathcal{P}:=\{f: \mathbb{R} \longrightarrow \mathbb{C} \; ; \; f \: \text{is infinitely differentiable and periodic with period}\: L\}. $$

For $r \in \mathbb{N}$ consider the Sobolev Space $H^r_{per}([0,L])$. This space can be interpreted as the set of $f \in \mathcal{P}'$ such that $$f, f' , \cdots, f^{(r)} \in L^2_{per}([0,L]),$$ with norm $$||f||_{H^r_{per}}=\left(\sum_{i=0}^{r}||f^{(i)}||^2_{L^2_{per}} \right)^{\frac{1}{2}}, \; \forall \; f \in H^r_{per}([0,L]), \tag{1}$$ where $$L^2_{per}([0,L])=\{f: \mathbb{R} \longrightarrow \mathbb{C} \; ; \; f \: \text{is periodic with period} \; L \; \text{and} \; f|_{[0,L]} \in L^2([0,L])\} $$ with inner product $$(f,g)_{L^2_{per}}=\int_0^L f(x) \overline{g(x)}\; dx, \; \forall \; f,g \in L^2_{per}([0,L]).$$

Now, consider $$ L^2_{per,m}([0,L]):=\left\{ f \in L^2_{per}([0,L]) \; \; \; \int_0^L f(x) \; dx=0\right\}. $$ Clearly, $L^2_{per,m}([0,L])$ is closed subspace of $L^2_{per}([0,L])$. From the Proposition $3.193$. of $[1]$ we have, for $r,s \in \mathbb{N}, s \geq r$, that $H^s_{per}([0,L])$ is continuously and densely embedded in $H^r_{per}([0,L])$. Simbolically $H^s_{per}([0,L]) \hookrightarrow H^r_{per}([0,L])$.

Question. Is $H^s_{per}([0,L]) \cap L^2_{per,m}([0,L]) \hookrightarrow H^r_{per}([0,L]) \cap L^2_{per,m}([0,L])$ also?

I think that we can write $$ H^n_{per,m}([0,L]) := H^n_{per}([0,L]) \cap L^2_{per,m}([0,L])=\left\{ f \in H^n_{per}([0,L]) \; \int_0^L f \; dx=0\right\}. $$ In this way, I think that $H^n_{per,m}([0,L])$ is closed subspace of the $H^n_{per}([0,L])$, and so is Hilbert space (with the induced inner product) for all $n\in \mathbb{N}$. In this sense, I think we would have $H^s_{per,m}([0,L])$ is continuously embedded in $H^r_{per,m}([0,L])$ (with the induced inner product). But is dense?

$[1]$ IORIO Jr., R. J.; IORIO, V. M. Fourier Analysis and Partial Differential Equations. Cambridge University Press, Cambridge, UK, 2001.