Convolution with imaginary Gaussian cannot be a $(p, p)$ operator unless $p=2$ 
Let $g=e^{-ix^2}, x\in \mathbb{R}$. Let $T$ be an operator defined as $T(f)=f*g$. Show that $T$ cannot satisfy a $(p,p)$ inequality unless $p=2$.
Note: We say an operator $T$ satisfies a $(p,p)$ inequality if there is a finite constant $C>0$ such that $\|Tf\|_p\le C\|f\|_p$ for all $f$.

An incomplete/flawed solution:
Fix $f\in L^P(\mathbb{R})$. Let $h(y)=f(y)e^{-iy^2}$. Then $h(y)\in L^p(\mathbb{R})$.
Observe that by some calculations we can arrive at
$$f*g(x)=e^{-ix^2}\hat{h}(-2x)$$
where $\hat{h}(x):=\int h(y)e^{-ixy}\,dy \ $ is the 'Fourier transform' of $h$.
Observe that $\hat{h} \in L^{p'}(\mathbb{R})$ where $\frac1{p'}+\frac1p=1$. Hence $f*g \in L^{p'}(\mathbb{R})$. Hence $p'=p$ implying $p=2$.
There are lots of mistakes in the above solution. For simplicity assume $1\le p\le 2$.  The Fourier transform of an arbitrary function $h$ in $L^p(\mathbb{R})$ is defined using the Riesz-Thorin interpolation. Is $\hat{h}$ defined above in the solution equal to the fourier transform of $h$? If we have proved $f*g \in L^{p'}(\mathbb{R})$ we need to construct a suitable $f$ such that $f*g \notin L^p(\mathbb{R})$, which seems difficult for me.
Can someone rectify or fill the holes in the above solution? Hints/suggestions are welcome.
 A: I think you're not on the right track. You already pointed out some of the problems with your approach. Let me try to help you see how to tackle such a problem.
First of all it helps to assume that $f$ is a Schwartz function. 
In fact a bit more than your claim is true. 

Claim. If for some $1\le p,q\le \infty$, there exists a constant $\infty>C>0$ such that for all Schwartz functions $f$,
  $$\|Tf\|_p \le C\|f\|_q.$$
  Then $p=q=2$.

Consider the function $g(x)=e^{-ix^2}$. It is an (imaginary) Gaussian function. It is merely a bounded function. However we can still talk about its Fourier transform, which is (a priori) a tempered distribution. One can show (e.g. using contour integration) that $\hat{f}$ is a bounded function, and
$$\hat{f}(\xi) = \int_{-\infty}^\infty e^{-ix^2} e^{-x\xi} dx = a e^{ib \xi^2},$$
where $a\in\mathbb{C}, b\in\mathbb{R}$ are constants whose exact values are not very interesting (they are $a=\frac12\sqrt{2\pi}(1-i)$ and $b=\frac14$). These values can be easily calculated by a purely formal computation.
To have fewer inessential constants jumping around let us set $a=1, b=-1/2$ and write $A\approx B$ and $A\lesssim B$ to mean $C^{-1} A\le B\le C A$ and $A\le C B$ with some finite, positive absolute constant $C$, respectively, where $A,B$ are positive quantities.
So $Tf(x)=\mathcal{F}^{-1} ( \hat{f}(\xi) e^{-i\xi^2/2} )(x),$
where $\mathcal{F}^{-1}f(x)=\int e^{ix\xi} f(\xi) d\xi$ is the inverse Fourier transform.
To get a feeling for such inequalities it is good to plug in a rescaled bump function and then see how both sides of the inequality behave when the scaling is changed. Since $g$ is a Gaussian, it is natural to plug in rescaled Gaussian's for $f$. Let $\lambda>0$ and set
$$f(x)=e^{-(x/\lambda)^2/2}.$$
We have $\|f\|_q\approx \lambda^{1/q}$.
We can compute $\hat{f}(\xi) = c \lambda e^{-(\lambda\xi)^2/2}$, where $c=\sqrt{2\pi}$. Then
$$ \hat{f}(\xi) e^{-i\xi^2/2} = c\lambda e^{-\frac{\xi^2}2 (\lambda^2+i)}$$
Using $\lambda^2+i=(e^{\frac12 \log(\lambda^2+i)})^2$ go ahead and compute 
$$ |\mathcal{F}^{-1}( \hat{f}(\xi)e^{-i\xi^2/2} )(x)| \approx \lambda |\lambda^2+i|^{-1/2} e^{-\frac{x^2}2 \text{Re}\Big(\frac1{\lambda^2+i}\Big)}=\lambda (\lambda^4+1)^{-1/4} e^{-\frac{x^2}2 \frac{\lambda^2}{\lambda^4+1}}$$
Thus,
$$ \|Tf\|_p \approx \lambda (\lambda^4+1)^{-1/4} \Big(\int e^{-\frac{x^2}2 \frac{p\lambda^2}{\lambda^4+1}} dx\Big)^{1/p}
\approx \lambda^{1-\frac1p} (\lambda^4+1)^{\frac1{2p}-\frac14}
$$
If we want $\|Tf\|_p\lesssim \|f\|_q$ to hold, then it is necessary that 
$$\lambda^{1-\frac1p-\frac1q} (\lambda^4+1)^{\frac1{2p}-\frac14}\lesssim 1$$
for all $\lambda>0$. But then we need to have $p=q=2$ (check it).
Note that $\|Tf\|_2\lesssim \|f\|_2$ really holds by Plancherel's theorem.
