# Relation between Fourier transform and uncertainty principle

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!

• 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 $f$. Commented Dec 5, 2021 at 6:56

$$\begin{split} \left|\widehat{f}(\lambda) -1\right| &= \left|\int_{-\infty}^\infty f(x)e^{-i\lambda x} dx -\int_{-\infty}^\infty f(x) dx\right|\\ &= \left| \int_{-\delta}^{\delta} f(x)\left(e^{-i\lambda x}-1\right)dx\right|\\ &\leq \int_{-\delta}^\delta f(x)|e^{-i\lambda x} -1|dx \\ &\leq 2\int_{-\delta}^{\delta}f(x)\left|\sin\left(\frac \lambda 2 x\right)\right|dx \end{split}$$ 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 $$\widehat{f}(\lambda) \geq 1-\frac 1 {100}=.99 \geq \frac 1 2$$

• Wow, thank you! Commented Dec 6, 2021 at 1:23

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 $$| \hat f |^2$$.