Considering the function $f:\mathbb{R} \to \mathbb{C}$, with $\left| f(x) \right|=1$ for all $x\in \mathbb{R}$.
Considering $g:\mathbb{R} \to \mathbb{C}$ with $\int_{-\infty}^{\infty}{\left|g(x)\right|^2dx}=1$
I am interested on properties of the amplitude of the Fourier Transform of the product of $f$ and $g$:
$A(k)=\left|FT(f(x)g(x))\right|$
Is there any constraint on $A$ apart from the fact that A is real positive? Considering a fixed $g$, is it possible to attain any $A$ simply by changing $f$?
Thank you for any help
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2 Answers
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If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\|g\|_{L^1}$.
Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof. (Except for special case: if $g$ is in $L^1$, then we can chop it into pieces with disjoint supports and small $L^1$ norm, and then use $f$ to move the Fourier transforms of pieces far from one another.)
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$\begingroup$ I have a special case of this lower bound problem: More precisely, let f(x)=\exp(i \theta(x)) and assume |\theta'(t)|<= \Omega and f(0)=f(a)=1. Define [the windowed FT] A(k)=\int_0^a \exp(i k t) f(x) dx How does a lower bound for [L2 norm on a fixed interval] \int_0^k_c |A(k)|^2 dk scale with \Omega? I suspect that the lower bound should be O(1/\Omega^2) or something similar but I am often wrong. Maybe I should pose this as a separate question? $\endgroup$ Commented Feb 23, 2010 at 23:05
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The only thing you can say is $\int_{-\infty}^{+\infty}|A(k)|^2dk=1$, since the Fourier transform is an isometry of $L^2(\mathbb{R})$.
Any function $A(k)$ with $\int_{-\infty}^{+\infty}|A(k)|^2dk=1$ is the Fourier transform of a function of the form $f\cdot g$. Let $h$ be the inverse Fourier transform of $A(k)$, so that $A=\hat h$ and $\int_{-\infty}^{+\infty}|h(x)|^2dx=1$. We can write $h=g\cdot f$ with $g=h/f$; then $$ \int_{-\infty}^{+\infty}|g(x)|^2dx=\int_{-\infty}^{+\infty}|h(x)|^2dx=1 $$ and $A=\widehat{f\cdot g}$
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$\begingroup$ I misunderstood the question. The size of the Fourier transform of g at infinity is "determined" by the smoothness of g. Multiplying by f does not change the smoothness, so I would not expect a change in the size of |A(k)|. $\endgroup$ Commented Feb 10, 2010 at 16:14
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$\begingroup$ If we add the hypothesis that everything is $C^\infty$-smooth, including f, does it help? And how would you go to find the right f for given functions A and g? Thank you a lot for your help $\endgroup$ Commented Feb 11, 2010 at 10:01