2
$\begingroup$

Let $f \in L^1 \cap C$ such that $f(x) \geq 0$ and $$\int_{-\infty}^\infty f(x) = 1$$ and such that $f(x) = 0$ for $|x| > \delta >0$.

Show that $|\widehat{f}(\lambda)|\geq \frac{1}{2}$ for $|\lambda| \leq \frac{1}{100\delta}$ where $$\widehat{f}(\lambda) = \int_{-\infty}^\infty f(x)e^{-i\lambda x} dx$$ This question shows that the smaller the support of a function, the more spread out the Fourier transform is.

I'm having trouble with the fact that I can't express the fourier transform explicitely and how to relate the fact that $||f||_1 = 1$ and $f(x)\geq 0$ for all $x$ and $f(x) = 0$ for $|x|>\delta$.

I was thinking that it had something to do with the inversion theorem but I can't seem to find any way to use it with those information.

Any hint or help would be very appreciated!

$\endgroup$
1
  • $\begingroup$ Probably one should use that the Fourier transform is 1 at the origin and then use some uniform continuity (boundedness of the derivative) of the Fourier transform, which comes from the support condition on $f$. $\endgroup$
    – PhoemueX
    Commented Dec 5, 2021 at 6:56

2 Answers 2

1
$\begingroup$

$$\begin{split} \left|\widehat{f}(\lambda) -1\right| &= \left|\int_{-\infty}^\infty f(x)e^{-i\lambda x} dx -\int_{-\infty}^\infty f(x) dx\right|\\ &= \left| \int_{-\delta}^{\delta} f(x)\left(e^{-i\lambda x}-1\right)dx\right|\\ &\leq \int_{-\delta}^\delta f(x)|e^{-i\lambda x} -1|dx \\ &\leq 2\int_{-\delta}^{\delta}f(x)\left|\sin\left(\frac \lambda 2 x\right)\right|dx \end{split}$$ If $|\lambda| \leq \frac{1}{100\delta}$, then for any $x\in[-\delta, \delta]$, we have $\left|\sin\left(\frac \lambda 2 x\right)\right|\leq \left|\frac \lambda 2 x\right|\leq \frac 1 {200}$. Thus $$\left|\widehat{f}(\lambda) -1\right|\leq \frac 1 {100}\int_{-\delta}^{\delta}f(x)dx=\frac 1 {100}$$ This implies that $$\widehat{f}(\lambda) \geq 1-\frac 1 {100}=.99 \geq \frac 1 2$$

$\endgroup$
1
  • $\begingroup$ Wow, thank you! $\endgroup$
    – Ianatore
    Commented Dec 6, 2021 at 1:23
1
$\begingroup$

Hints. Note that the real function $f(x)$ can be regarded as a probability density.

  1. Try expanding the Fourier transform $\hat f (\lambda)$ in its Taylor series, in powers of $\lambda$.

  2. The Taylor coefficients can be regarded as the moments of the probability distribution $f(x)$, and can be bounded by the associated powers of $\delta$. Also consider first the special case that $\delta <1$. Then these moments decay geometrically as powers of $\delta$

  3. It is then instructive to first focus on the special case that $f(x)$ is even. Then the series for $\hat f$ is also real and even.

Note that in this case the power series for $\hat f$ is an alternating series in even powers of $\lambda$, so its partial sums are alternating under-estimates and over-estimates of the true limit. The first two terms of that series should establish the desired inequality.

  1. In general, when $f$ is not assumed even, nevertheless the even part of $f$ corresponds to real part of $\hat f$, whose square provides a lower bound for the expression $| \hat f |^2$.
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .