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$f(x)$ is a 1D smooth function which real, even, compactly supported in the interval $[-a,a]$, and strictly positive within that interval.

Its Fourier transform is,

$$ \hat{f}(s) = \int_{-\infty}^{\infty} f(x) \exp(-2 \pi i x s) dx $$

Are the local minima of $|\hat{f}(s)|$ all zeros?

This is a follow up question to a previous question.

Filipe Maia
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2 Answers2

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It is not so easy to write down a function that satisfies your requirements and has an explicit formula for its Fourier transform. Luckily, an explicit form is not necessary:

Let $f$ be any function which is as specified in the question. Assume w.l.o.g. $\hat f(0) = 1$ and $0$ a global maximum of $|\hat f|$. Otherwise we can modulate and rescale $f$ to end up in this situation. Consider the function $g_a(x) := f(x) - f(x/a)/(2a)$. Note that $f, g_a$ are continuous and $\hat g_a= \hat f - \hat f(a x)/ 2$. Claim: $|\hat g_a|$ has a local minimum in a neighborhood of $0$ for some $a > 0$. Since $0$ is a global maximum of $|\hat f|$, there is a neighborhood $U$ of $0$ such that $|\hat g_a|>1/4$ for all $a>0$ and $|\hat f|>3/4$.

We clearly have that $|\hat g_a(0)| = 1/2$ for all $a>0$. Moreover, for $a \to \infty$ we see that $\hat g_a$ converges pointwise to $\hat f$ on $\mathbb R\setminus \{0\}$ and to $1/2$ on $\{0\}$. Thus, for sufficiently large $a$ we see that there are two points $x_1< 0 < x_2$ in $U$ such that $|\hat g_a(x_i)|>1/2 = |\hat g_a(0)|$ for $i = 1,2$. Thus $|\hat g_a|$ has a local minimum between $x_1$ and $x_2$ which is not a root of $\hat g_a$.

pcp
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  • The problem is that $g_a$ is no longer compactly supported in the interval $[−a,a]$. – Filipe Maia Jun 30 '18 at 17:07
  • I guess if one considers only $a < 1$ the first part works, but the argument starting with "for sufficiently large $a$" does not hold. – Filipe Maia Jun 30 '18 at 17:13
  • g_a is compactly supported by construction. It is not strictly positive on the full support though. I understood your question in the sense that the function should be strictly positive somewhere within the interval and not on the whole interval as the latter is impossible, since a function cannot be positive on all of its support. I suppose you meant strictly positive on (-a,a) then?. – pcp Jun 30 '18 at 18:33
  • But $g_a$ with $a > 1$ will spill outside the interval. – Filipe Maia Jun 30 '18 at 20:12
  • Yes, but as I've said, it has compact support. You can rescale the function to be supported in any interval you like, without changing the fact that the modulus of its fourier transform has a local minimum that is not a root. – pcp Jun 30 '18 at 20:44
  • Ah, I see what you mean now. I'll have to think about it a bit to make sure I understand the rest... – Filipe Maia Jun 30 '18 at 22:03
  • Well I think my answer is not going to be what you actually want, since the function will not be non-negative on its whole support. As I've said I probably misunderstood the question. – pcp Jun 30 '18 at 22:06
  • Even if it's not non-negative it might still give me some ideas. BTW you probably mean to say "root of $\hat g_a$" at the end of your answer. – Filipe Maia Jun 30 '18 at 22:13
  • Yes, you are right. I made some changes. – pcp Jun 30 '18 at 22:17
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Here's an example of a function whose Fourier transform has a local minima which is not a zero:

$$ f(x) = \begin{cases} 3(1 + \cos \pi x) + (1 + \cos 4 \pi x)& |x| \le 0.25, \\ 3(1 + \cos \pi x) & 0.25 < |x| < 1, \\ 0 & \text{otherwise} \end{cases} $$

Here's a plot of the function:

Plot of the function $f$

and of the absolute value of its Fourier transform:

Plot of the Fourier transform of $f$

Filipe Maia
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