Vtyjkyuk

I have shown that the inverse Fourier transform is unbounded from $L^p(mathbbR^d) to L^q(mathbbR^d)$ when $1le p <2$ ($p$ and $q$ being Holder conjugates). Here I define the Fourier transform
$$hatf(xi) = int_mathbbR^d e^-2pi i x cdot xi f(x) dx$$ and with no negative in the exponent for the inverse Fourier transform.
I was wondering whether we could use duality somehow to deduce that the Fourier transform is unbounded from $L^p(mathbbR^d$ to $L^q(mathbbR^d)$ for $p >2$.

The hope would be that we can use that the Fourier transform is unitary, i.e.,
$$int hatf barg = int f barcheckg.$$

But, this runs into problems when we consider formally what these operators are doing. By this I mean, with $mathcalF$ the Fourier transform, we have
$$mathcalF:L^p to L^q$$
$$mathcalF^-1:L^q to L^p$$
$$mathcalF^*: L^p to L^q$$

so while it's true that $mathcalF^-1$ and $mathcalF^*$ might agree on say, $mathcalS(mathbbR^d)$, the first equality below is not true:
$$|mathcalF^-1| = |mathcalF^*| = |mathcalF|$$
where those are operator norms, so we cannot immediately conclude unboundedness of the Fourier transform this way.

How should I proceed?

Edit: So, two things. The first is that I should've been saying "inverse of Fourier Transform" in everything above, not "inverse Fourier transform". The second thing is that I was able to adapt the proof I had to get that the Fourier transform is unbounded for $p>2$, so everything is good.