If Fourier transform of a simple function $f$ satisfies inequality $\| \hat{f} \|_q \leq C\|f\|_p$ then $1/p + 1/q = 1$

Question

I have some difficulties of getting started with this question. Namely,

If the Fourier transform $\hat f$ of a simple function $f$ satisfies $$\|\hat f \|_q \leq C\|f\|_p$$ then $1/q + 1/p = 1$.

Could someone give me a hint on how to get started?

Answer 1

Not sure if this is what you want.

If this claim is applied to all admissible $f$, then you can take $f = e^{-a x^2}$, then your inequality will give

$$\pi^{1 - \frac{1}{q} - \frac{1}{p}} q^{-1/q} \le C a^{1-\frac{1}{p} - \frac{1}{q}} p^{-1/p}. $$

which must be true for all $a$. Then changing $a\to 0$ and $a\to\infty$, which must force the power on $a$ as zero.

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Update. If you choose $f = \chi_{[-a/2, a/2]}$, then $\hat{f} = \text{sinc}(a \zeta)/\zeta$, you can derive if ($q > 1$).

$$a^{1 - \frac{1}{q} - \frac{1}{p}} \|\text{sinc}(x)\|_{L^q}\le C.$$

Then use the boundedness of the norm of sinc. Let $a\to 0$ and $a\to\infty$, you can get the same thing.