We know that there is no any inclusion relation between $L^p$ spaces on $\mathbb R^d.$ [e.g., Define $g(x)=\chi_{B}(x) |x|^{-a}$ where $B=\{x\in \mathbb R^d: |x|<1\}.$ Since $g$ is radial function, using polar coordinates, we have
$$\|g\|_{L^p}^p= C \int_{0}^1r^{-(1+ap-d)} dr<\infty$$
if $1+ap-d<1$. Choose $a=\frac{d-\epsilon}{p} $ (small $\epsilon>0$), then $g\in L^{p}(\mathbb R^d)$ and $g\notin L^q(\mathbb R^d)$ for $p<q.$ Thus we have $g\in L^p$ but not in $L^q.$ Similarly we may find $h\notin L^p$ but not in $h\in L^q.$]
We now define $L^p-$Sobolev spaces by the norm $$\|f\|_{L^p_s} = \| \mathcal{F}^{-1} \left(\langle \cdot \rangle^s \mathcal{F}f(\cdot)\right)\|_{L^p} $$ where $\langle \cdot \rangle = (1+ |\cdot|^2)^{1/2}, s\in \mathbb R$ and $\mathcal{F}$ and $\mathcal{F}^{-1}$ denotes the Fourier transform and inverse Fourier transform respectively.
Note: when $p=2,$ usually $L^p_s$ is denoted by $H^s$
Question: What is an inclusion relation between $L_s^p-$Sobolev spaces? Can we find $f\in L^p_s(\mathbb R^d)$ but not in $L^q_s(\mathbb R^d)$ for $p\neq q$. ($1\leq p, q \leq \infty$)
The answer to our question is yes. Sketch of the proof: Fix $p \neq q$ and let $f \in L^p(\mathbb{R}^n)$ and $g \in L^q(\mathbb{R}^n)$ with $f \neq g$. You can define (using distribution theory) an injective map from $J: L^l(\mathbb{R}^n) \rightarrow L^l_s(\mathbb{R}^n)$ given by $$ J(f) = \mathcal{F}^{-1} \big(\langle \cdot \rangle^{-s} \, \mathcal{F}(f)\big). $$ But now, $J(f) \in L_s^p(\mathbb{R}^n)$, $J(g) \in L_s^q(\mathbb{R}^n)$ and $f \neq g$.