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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.

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Not sure if this is true, but me may write that

\begin{eqnarray} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} &=& \frac{\tau+|\xi|^4+i}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+i} \\ &:=& m_{\varepsilon}(\tau, \xi) \hat{F}(\tau, \xi) \end{eqnarray}

and from the case $\varepsilon = 0$, you already know that $F$ belongs to $L^{p'}$. Then, shouldn't the desired conclusion follow from the Mikhlin multiplier theorem applied to the multiplier $m_{\varepsilon}$ ?

Update : Following your suggestion of computing the inverse Fourier transform, the operator $T_{\varepsilon}$ associated to your fraction may be written informally as

$$T_{\varepsilon}(t, \xi) = e^{it \varepsilon|\xi|^2}T_0(t, \xi)$$

with $T_0$ continuous from $L^p$ to $L^{p'}$. So your result amounts now to prove that the standard Schrödinger kernel defines a continuous operator on $L^{p'}$ uniformly in $\varepsilon$, which sounds reasonable.

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  • $\begingroup$ Thanks! I'm mainly concerned with the uniformity of the bound, and it seems that the Mikhlin multiplier theorem also implies that too. $\endgroup$
    – Tomas
    Commented Jan 30, 2015 at 9:49
  • $\begingroup$ Probably. I suggest that you have a close look at the proof. Because the theorem only assumes some decay on the multiplier, one may expect that the implied continuity constant only depend on this decay, which is here uniform in $\varepsilon$. $\endgroup$
    – Hachino
    Commented Jan 30, 2015 at 9:52
  • $\begingroup$ I think I was wrong, just take one derivative for $\tau$, we don't have a decay like $\frac{C}{\sqrt{\tau^2+|\xi|^2}}$. $\endgroup$
    – Tomas
    Commented Jan 30, 2015 at 11:22
  • $\begingroup$ The derivative of $m_{\varepsilon}$ with respect to $\tau$ is equal to $\frac{\varepsilon |\xi|^2}{(\tau + |\xi|^4 + \varepsilon |\xi|^2 + i)^2}$, whose modulus behaves like $\frac{1}{(1+\tau)^2}$ both at $0$ and infinity. So decay is even faster than what you need. $\endgroup$
    – Hachino
    Commented Jan 30, 2015 at 11:38
  • $\begingroup$ I was a little bit confused. Since we are working in $\mathbb{R}^{n+1}$, shouldn't the bound be the form I just wrote? Note also that $\tau, \epsilon \in\mathbb{R}$ can be arbitrary, I still don't think that we can use Mikhlin's theorem here. $\endgroup$
    – Tomas
    Commented Jan 30, 2015 at 11:54

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