Let $\Omega\subset\mathbb{R}^n$ be a bounded smooth domain and $n>1$ and let $T_p\colon W^{1,p}(\Omega)\to L^p(\partial\Omega)$ denote the trace theorem for $p\in[1,\infty]$.
Suppose, we are given $u\in W^{1,2}(\Omega)\cap L^q(\Omega)$ for some $q\in(2,\infty]$. Of course, we have $T_2u\in L^2(\partial\Omega)$ with the estimate
$$ \|T_2u\|_{L^2(\partial\Omega)}\leq C_p\left(\|u\|_{L^2(\Omega)}+\|\nabla u\|_{L^2(\Omega)^n}\right) $$
where the constant $C_p$ denotes the continuity constant of $T_p$.
My question: is there any way to exploit the higher integrability of $u$ to get higher integrability of the trace, e.g., something like
$$ \|T_ru\|_{L^r(\partial\Omega)}\leq \tilde{C}\left(\|u\|_{L^s(\Omega)}+\|\nabla u\|_{L^2(\Omega)^n}\right) $$
for any $r,s\geq 2$ where $r>2$ and where $\tilde{C}$ might be a different constant depending on $r,s$ but still uniform for every $u$?
Consider the case of the half space $\Omega = \mathbb R^n_+ = \{ x \in \mathbb R^n : x_n > 0\}$. The general case follows by a suitable flattening + patching argument, which I omit the details of.
Suppose $u \in W^{1,2}\cap L^q$, and observe that for $1<s<\infty$ to be determined we have \begin{align*}\int_{\mathbb R^{n-1}} \lvert u(x',0) \rvert^s \,\mathrm{d}x' &= - \int_0^{\infty} \int_{\mathbb R^{n-1}} \partial_{x_n}(\lvert u(x',x_n) \rvert^s) \,\mathrm{d}x'\,\mathrm{d}x_n \\ &\leq s \int_{\mathbb R^n_+} \lvert u \rvert^{s-1} \lvert \mathrm{D}u\rvert \,\mathrm{d}x \\ &\leq s \left(\int_{\mathbb R^n_+} \lvert u \rvert^{\frac{p(s-1)}{p-1}}\,\mathrm{d}x \right)^{\frac{p-1}p} \left(\int_{\mathbb R^n_+} \lvert \mathrm{D}u \rvert^{p}\,\mathrm{d}x \right)^{\frac1p}.\end{align*} Here we have (modulo some technicalities) used the fundamental theorem of calculus in the first line (in the $x_n$-variable), and Hölder in the last line. Up to here is analogous to how one usually proves the trace theorem, such as in Evans' PDE text (Section 5.5), except that one normally takes $s=p$.
In our case we set $p=2$, and since $u \in L^q$ we can choose $s>1$ such that $$ q = \frac{p(s-1)}{p-1} = 2(s-1) \iff s = \frac{q}2 + 1.$$ Thus we obtain an improvement in integrability.
Note that by Sobolev embedding we necessarily can take $q = \frac{2n}{n-2}$, so in this case we get $s = \frac{2(n-1)}{n-2}$ (and we can take any $s<\infty$ if $n=2$). Also as a side-note, we do not get any improvement this way if $p=1$, regardless of how large $q$ is.