Relation between Fourier transform and uncertainty principle
Solution 1:
$$\begin{split} \left|\widehat{f}(\lambda) -1\right| &= \left|\int_{-\infty}^\infty f(x)e^{-i\lambda x} dx -\int_{-\infty}^\infty f(x) dx\right|\\ &= \left| \int_{-\delta}^{\delta} f(x)\left(e^{-i\lambda x}-1\right)dx\right|\\ &\leq \int_{-\delta}^\delta f(x)|e^{-i\lambda x} -1|dx \\ &\leq 2\int_{-\delta}^{\delta}f(x)\left|\sin\left(\frac \lambda 2 x\right)\right|dx \end{split}$$ 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 $$\widehat{f}(\lambda) \geq 1-\frac 1 {100}=.99 \geq \frac 1 2$$
Solution 2:
Hints. Note that the real function $f(x)$ can be regarded as a probability density.
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Try expanding the Fourier transform $\hat f (\lambda)$ in its Taylor series, in powers of $\lambda$.
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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 $\delta$
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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.
- 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 $| \hat f |^2$.