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Choose a nonzero $f\in L^p({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ whose transform belongs to $L^q$, for example a Gaussian by the Fourier transform of a Gaussian. For $\lambda >0$, set $f_{\lambda }(x)=f(\lambda x)$. The scaling property of the Fourier transform givesThe assumed estimate, applied for every $\lambda >0$, becomesSince both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the scaling necessity for an Lp to Lq Fourier bound yieldsHence $q$ is uniquely the conjugate exponent of $p$.