Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f$ is a strictly positive function in $L^1(\mathbb R)$. Show $$ |\hat f(y)| < \hat f(0) \text{, for all } y \ne 0. $$

Using monotonicity of the integral, I can show $|\hat f(y)| \le \hat f(0)$.
I don't see how to make the inequality strict.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

If $\widehat f(0)=|\widehat f(y)|$ for some $y\neq 0$, let $\theta$ such that $|\widehat f(y)|=e^{i\theta}\widehat f(y)$. Then $$\int_{\Bbb R}f(t)dt=\int_{\Bbb R}f(t)e^{i(ty+\theta})dt,$$ which gives $$\int_{\Bbb R}f(t)(1-e^{i(ty+\theta)})dt=0=\int_{\Bbb R}f(t)(1-\cos(ty+\theta))-i\int_{\Bbb R}f(t)\sin(ty+\theta)dt.$$ In particular $$\int_{\Bbb R}\underbrace{f(t)(1-\cos(ty+\theta))}_{\geq 0}dt=0,$$ so $f(t)(1-\cos(ty+\theta))=0$ almost everywhere. As $f>0$ almost everywhere, we get a contradiction.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.