Complex exponential is not Fourier multiplier on $L^p$ I am having difficulty to show that the function $m(\xi):= e^{i|\xi|^2}$ is not a Fourier multiplier on $L^p$ when $p\neq 2$.
Note that $m:\mathbb{R}^n\to \mathbb{C}$ is called an $L^p$ Fourier multiplier if
$$\|T_m f\|_p := \|(m\hat{f})^\vee\|_p \leq C\|f\|_p$$
for some $C>0$, where $\hat{f}$ is the Fourier transform of $f$, and $f^\vee$ is its inverse Fourier transform.
Please help, thank you.
 A: Intuition: this multiplier evolves $f$ by the Schrodinger equation by time one (after making Planck's constant one).  And this maps the Dirac delta function to something whose absolute value is constant, which is a counterexample for $p=1$.  So approximate the Dirac delta function by a narrow bell curve.  This will work for $p<2$.  For $p>2$, do the same argument backwards in time (i.e. what maps onto the Dirac delta function).
If $p < 2$, consider $f(x) = e^{-\frac1{2a} |x|^2}$ for $a>0$.  Then $T_mf(x) = \left(\frac a{a+4i}\right)^{\frac n2} e^{-\frac1{2a+4i} |x|^2}$, that is, $|T_mf(x)| = \left(\frac{a^2}{a^2+256}\right)^{\frac n4} e^{-\frac{a}{2a^2 + 8} |x|^2}$.  Thus if $a \ll 1$
$$ \|f\|_p \sim a^{\frac n{2p}} ,$$
$$ \|Tf\|_p \sim a^{\frac n2 -\frac n{2p}} .$$
Letting $a \to 0$, we see that $\|T_m\|_p / \|f\|_p \sim a^{n\left(\frac12-\frac1p\right)}\to \infty$.
If $p > 2$, consider $f(x) = e^{-\frac1{2a-4i} |x|^2}$ for $a>0$.  Then $T_mf(x) = \left(\frac a{a-4i}\right)^{\frac n2}e^{-\frac1{2a} |x|^2}$.  Argue as before.
Signs and constants might differ, depending upon which definition you use for the Fourier transform.  Details may be wrong, but the idea should work.
A: I'd like to give a proof without given a precise example. But we need to use the Riesz-Thorin interpolation theorem and the following lemma.
Lemma: The inverse Fourier transform of $e^{i\pi|\xi|^2}$ is $e^{i\frac{\pi n}{4}}e^{-i\pi |\xi|^2}$, a bounded function.
Proof: Suppose that $e^{i|\xi|^2}$ is a $L^p$ multiplier, after a dilation, $e^{i\pi|\xi|^2}$ should also be a $L^p$ multiplier.
Define $Tf= (\hat{f}(\xi)e^{i\pi|\xi|^2})^{\vee}=f* (e^{i\pi|\xi|^2})^{\vee}$, then for any $f\in L^1$,
$$
\|Tf\|_{L^\infty}\le \int |f(y)| dy \|e^{i\frac{\pi n}{4}}e^{-i\pi |\xi|^2}\|_{L^\infty}= \|f\|_{L^1}.
$$
And, by Plancherel's inequality, we have
$$
\|Tf\|_2^2=\| \widehat{Tf}\|_2^2 =\|\hat{f}\|_2^2=\|f\|_2^2.
$$
These two facts imply that $T$ is maps $L^1$ to $L^\infty$ and $L^2$ to $L^2$. By Riesz-Thorin interpolation theorem, we deduce that
$$
\|Tf\|_{p'}\le \|f\|_p.
$$
Now suppose that $T$ is also a $L^p$ multiplier, we then have
$$
\| Tf\|_p\le \|f\|_p.
$$
If $1\le p<2$, use Riesz-Thorin's theorem again, we have
$$
\|f\|_2^2=\|Tf\|_2\le \|f\|_p,~~~~~~~~\forall f\in L^p.
$$
This implies that $L^p\subset L^2$. This is a contradiction. Since a $L^p$ multiplies is also a $L^{p'}$ multiplier, when $p\not=2$, it can not be a $L^p$ multiplier.
