Complex exponential is not Fourier multiplier on $L^p$

Question

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.

Answer 1

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.

Answer 2

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.