Fourier transform on functions on the unit sphere bounded by the Fourier transform of measure

Question

Let $f$ be a non-negative simple function on the unit sphere. So that the absolute value of the Fourier transform of $f(y)$ on the unit sphere is $$|\widehat{fd\sigma}| = |\int_{\mathbf{S}^n}\sum_{i=1}^N\alpha_i1_{A_i}e^{-i2\pi\langle x, y\rangle} d\sigma(x)| = |\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}|.$$ Now, isn't that true if we have $0 < \alpha_i \leq 1,$ for all $i \in \{1,...,N\}$ then we have $$|\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}| \leq |\sum_{i=1}^N\widehat{1_{A_i}d\sigma}| = |\widehat{d\sigma}|?$$ EDIT: Then we would have for general bounded $f = f_1^+ - f_1^- +if_2^+ - if_2^-$ that $$ ||\widehat{fd\sigma}||_q \leq 4\max_{f_1^+,f_1^-,f_2^+,f_2^-}||\widehat{f_i^j\sigma}||_q \leq 4||f||_{\infty}||\widehat{d\sigma}||_q$$ via Minkowski and a standard limit process.

Answer 1

As the Fourier transform on the sphere is real valued, it follows that $$|\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}(y)| =|\alpha\widehat{1_{A_1}d\sigma(y)} + \beta\widehat{1_{A_2}d\sigma(y)}|,$$ for some $A,B$ with $A \cup B = \mathbf{S}^n$ and for some $0 \leq \alpha,\beta \leq 1.$ Above $$A_1 := \{ x \in \mathbf{S}^n | \cos{\langle x, y \rangle} \geq 0 \}$$ and $$A_2 := \{ x \in \mathbf{S}^N | \cos{\langle x,y \rangle} < 0 \}.$$ Now, there at least exist a point wise maximizer because $$|\alpha\widehat{1_{A_1}d\sigma(y)} + \beta\widehat{1_{A_2}d\sigma(y)}| \leq \max_{A_1,A_2}|\widehat{1_{A_i}d\sigma(y)}| := g(y)$$ This means that for all bounded positive functions $f$ it holds that $|\widehat{fd\sigma}(y)|$ is bounded point wise by $||f||_{\infty}g(y).$ However, the above function $g$ is not dominated point wise by $|\widehat{d\sigma}|.$