Let $(\tau,\xi)\in\mathbb{R}\times \mathbb{R}^n$, and consider the function $$ m_{\epsilon}(\tau,\xi)=\frac{1}{\tau+|\xi|^4+\epsilon|\xi|^2+i} $$ in $\mathbb{R}^{n+1}$. My first question is that does the function $m_{\epsilon}(\tau,\xi)$ define a $L^p(\mathbb{R}^{n+1})-L^{p'}(\mathbb{R}^{n+1})$ multiplier with $p=\frac{2(n+4)}{n+8}$, $p'=\frac{2(n+4)}{n}$ for all $\epsilon\in\mathbb{R}$? Or equivalently, do we have the following estimates $$ \|\mathcal{F^{-1}}(\frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+\epsilon|\xi|^2+i})\|_{L^{p'}(\mathbb{R}^{n+1})}\leq C\|f\|_{L^p(\mathbb{R}^{n+1})}~~~~ ? $$ Where $\mathcal{F^{-1}}$ denotes the Fourier inversion in $\mathbb{R}^{n+1}$..

If $\epsilon=0$, this is true, which is implied by the fact that $$ \|e^{it{\Delta^2}}u\|_{L^{p'}(\mathbb{R}^n)}\leq C|t|^{-\frac{n}{n+4}}\|u\|_ {L^{p}(\mathbb{R}^n)} $$ and 1-d H-L-S inequality. I'm interested in the case where $\epsilon\ne 0$, now the symbol is no longer homogeneous, so the above bound can't be derived directly. In particular, does the following uniform estimates $$ \|e^{it({\Delta^2+\epsilon\Delta)}}u\|_{L^{p'}(\mathbb{R}^n)}\leq C|t|^{-\frac{n}{n+4}}\|u\|_ {L^{p}(\mathbb{R}^n)} $$ hold with C independent of $\epsilon$?

Thanks for any comments or references.