Related to Hausdorff-young inequality: p,q>=1. If exists C, $\|\hat f\|_{q}\le C\| f\|_{p}$ for all $f\in L^p(\mathbb R^n)$, is 1/p+1/q=1? Hausdorff-young inequality: If $1\le p\le2, f\in\mathbb L^p(\mathbb R^n)$, then $\|\hat f\|_{q}\le \|  f\|_{p}$, where $1/p+1/q=1$.  
Here is a question:
If $1\le p,q\le\infty$ (that is, p can be bigger than 2) there exists a constant $C$ such that $$\|\hat f\|_{L^{q}(\mathbb R^n)}\le C\|  f\|_{L^{p}(\mathbb R^n)}$$ for all $$f\in L^p(\mathbb R^n),$$ can you conclude that $\frac1p+\frac1q=1$?
 A: This is indeed true. Let $f$ be a non zero function and define for any $\lambda >0$:
$$f_\lambda(x)=f(\lambda x)$$
Then with the change of variables $y=\lambda x$:
$$\|f_\lambda\|_{L^p}=\left(\int_{\mathbb{R}^n} |f(\lambda x)|^p dx \right)^\frac{1}{p}=\left(\lambda^{-n} \int_{\mathbb{R}^n} |f(y)|^p dy\right)^\frac{1}{p}=\lambda^\frac{-n}{p}\|f\|_{L^p}$$
(and $\|f_\lambda\|_{L^\infty}=\lambda^0 \|f\|_{L^\infty}=\lambda^\frac{-n}{\infty} \|f\|_{L^\infty}$)
More over using the same change of variables:
$$\widehat{f_\lambda}(\xi)=\lambda^{-n} \widehat{f} \left(\frac{\xi}{\lambda} \right)$$
so:
$$\|\widehat{f_\lambda}\|_{L^q}=\lambda^{-n} \left( \frac{1}{\lambda} \right)^\frac{-n}{q} \|\widehat{f}\|_{L^q}$$
so the inequality apllied on $f_\lambda$ becomes:
$$\lambda^{-n\left(1+\frac{1}{q} \right)} \|\widehat{f}\|_{L^q} \leq \lambda^{-n \frac{1}{p}} \|f\|_{L^p}$$.
Taking $\lambda \to +\infty$ you see that you must have:
$$-n \frac{1}{p} \geq -n\left(1+\frac{1}{q} \right)$$
and taking $\lambda \to 0$ you must also have:
$$-n \frac{1}{p} \leq -n\left(1+\frac{1}{q} \right)$$
so:
$$\frac{1}{p}+\frac{1}{q}=1$$
