$m(\xi)=(-1)^{[\xi^2]}$ is not a Fourier multiplier 
Let
$$m(\xi)=(-1)^{[\xi^2]},\qquad \xi\in\mathbb R,$$
where $[a]$ means the largest integer less than or equal to $a$. Prove that $m$ is not a Fourier multiplier on $L^p(\mathbb R)$ for some $p\neq 2$.

Note that $m:\mathbb{R}\to \mathbb{C}$ is called an $L^p(\mathbb R)$ 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:
$$\hat f(\xi)=\int_\mathbb{R} f(x)e^{-2\pi ix\xi}\,dx,\qquad f^\vee(x)=\int_\mathbb{R} f(\xi)e^{2\pi ix\xi}\,d\xi.$$
The problem comes from the following theorem:
Theorem. If $m\in C(\mathbb R)$ is bouneded such that $m^2$ has bounded variation, then $m$ is an $L^p(\mathbb R)$ Fourier multiplier.
I was wondering whether the condition $m\in C(\mathbb R)$ is redundant, and my teacher told me this counterexample. But I don't know how to show the counterexpamle.
My try. I tried the functions $f_N$ such that $\hat{f_N}=\chi_{[0,\sqrt N]}$. I wish to show that
$$\sup_{N}\frac{\|T_m f_N\|_p}{\|f_N\|_p}=\infty.$$
It is direct to calculate that $f_N(x)=\frac{e^{2\pi i\sqrt N\ x}\ -1}{2\pi ix}$, so $\|f_N\|_p\sim N^{\frac12-\frac1{2p}}.$ But I don't know how to estimate $\|T_m f_N\|p$, so I stuck.
Any help would be appreciated!
 A: My teacher gave me some hints and I finally figure out the proof. The hints are given in the following picture. Although there are some Chinese characters, I think it is not hard to understand the math. I think it may be better for someone to think first with the hints before reading my answer. Not every sentence in the hints is obvious.

Before we start our proof, we state a lemma which is useful for getting the lower bound.$\newcommand{\RR}{\mathbb R}$

Lemma. Suppose that $\varphi(x)$ is a bounded measurable periodic function on $\RR$ with period $T>0$, and $f\in L^1(\RR)$. Then
$$\lim_{|\lambda|\to+\infty}\int_\RR f(x)\varphi(\lambda x)\,dx=\left(\frac1T\int_0^T\varphi(x)\,dx\right)\int_\RR f(x)\,dx.$$

Now we prove that $m(\xi)=(-1)^{[\xi^2]}$ is not a $L^p$ multiplier for any $p\neq 2$.
For each $a>0$, let $f(x)=a^{-1/2}e^{-\frac{\pi x^2}{a}}$, $g(x)=(a+i)^{-1/2}e^{-\frac{\pi x^2}{a+i}}$, $x\in\RR$. Then $\hat f(\xi)= e^{-\pi a\xi^2}, \hat g(\xi)=e^{-\pi(a+i)\xi^2}, \xi\in\RR$. Assume that $m$ is an $L^p(\RR)$ multiplier, then we have
$$\left|\int_\RR m(\xi)\hat f(\xi)\hat g(\xi)\,d\xi\right|=\left|\int_\RR(T_mf)(x)g(-x)\,dx\right|\leq C_p \|f\|_p\|g\|_q, \tag{1}$$
where $q$ is the conjugate exponent of $p$.
By our choice of $f$ and $g$,
$$\|f\|_p\sim a^{\frac1{2p}-\frac12},\qquad \|g\|_q\sim a^{-\frac1{2q}}(a^2+1)^{\frac1{2q}-\frac14}. \tag{2}$$
We also have
\begin{align*}
\left|\int_\RR m(\xi)\hat f(\xi)\hat g(\xi)\,d\xi\right|&=\left|\int_\RR(-1)^{[\xi^2]}e^{-2\pi a\xi^2}(\cos(\pi\xi^2)+i\sin(\pi\xi^2))\,d\xi\right|\\
&\geq\left|\int_\RR(-1)^{[\xi^2]}e^{-2\pi a\xi^2}\sin(\pi\xi^2)\,d\xi\right|\\
&=\int_\RR e^{-2\pi a\xi^2}|\sin(\pi\xi^2)|\,d\xi\\
&\geq \int_{|\xi|\geq 1}e^{-2\pi a\xi^2}|\sin(\pi\xi^2)|\,d\xi\\
&=a^{-\frac12}\int_{t\geq 1}\frac{e^{-2\pi t}}{\sqrt t}\left|\sin\left(\frac{\pi t}a\right)\right|\,dt.
\end{align*}
Let
$$\varphi(t)=|\sin(\pi t)|,\qquad F(t)=\begin{cases}
 \frac{e^{-2\pi t}}{\sqrt t}, & t\geq 1,\\
 0, & t<1.
\end{cases}$$
Then $\varphi$ is a bounded measurable function with period $1>0$ and $F\in L^1(\RR)$. By our Lemma,
$$\lim_{a\to0^+}\int_{t\geq 1}\frac{e^{-2\pi t}}{\sqrt t}\left|\sin\left(\frac{\pi t}a\right)\right|\,dt=\left(\int_0^1\varphi(t)\,dt\right)\int_\RR F(t)\,dt=:2K>0.$$
So, there exists $\delta>0$ such that
$$\left|\int_\RR m(\xi)\hat f(\xi)\hat g(\xi)\,d\xi\right|\geq Ka^{-\frac12},\qquad a\in(0,\delta).  \tag{3}$$
Combining $(1)$, $(2)$ and $(3)$ gives that
$$Ka^{-\frac12}\leq C a^{\frac1{2p}-\frac12}\cdot a^{-\frac1{2q}}(a^2+1)^{\frac1{2q}-\frac14}, \qquad a\in(0,\delta).$$
Using $\frac1p+\frac1q=1$, we obtain
$$a^{\frac1q-\frac12}(a^2+1)^{\frac14-\frac1{2q}}\leq C, \qquad a\in(0,\delta),$$
which implies that $\frac1q\geq\frac12$, i.e., $q\leq 2$, or $p\geq 2$.
We have proved that: if $m$ is an $L^p$ multiplier, then $p\geq2$. Assume that $m$ is an $L^p$ multiplier for some $p>2$. Let $q$ be the conjugate exponent of $p$, then $q<2$, and
\begin{align*}
\|T_m f\|_q&=\sup_{\|g\|_p\leq1}\left|\int_\RR T_mfg\,dx\right|=\sup_{\|g\|_p\leq1}\left|\int_\RR fT_mg\,dx\right|\\
&\leq \sup_{\|g\|_p\leq1}\|f\|_q\|T_mg\|_p \leq C_p\|f\|_q.
\end{align*}
So $m$ is an $L^q$ multiplier. Recall that $q<2$, which is a contradiction. So $m$ is an $L^p$ multiplier if and only if $p=2$.
Hints for the proof of Lemma:

 WLOG we assume that $\int_0^T\varphi(x)\,dx=0$. Firstly, prove the identity for $f=\chi_{[a,b]}$ and thus the identity holds for step functions $f$; finally, note that step functions are dense in $L^1$.

