Let $f,g \in L_2(\mathbb{R}) $, and let $\hat{f}$ and $\hat{g}$ denote their Fourier Transforms. I'm trying to find a lower bound for this sum:

$$ \left(\int_{-\infty}^{\infty} x^2 \, f^2(x) \, dx \int_{-\infty}^{\infty} k^2 \, \widehat{g}^2(k) \, dk \right)+ \left(\int_{-\infty}^{\infty} k^2 \, \hat{f}^2(k) \, dk \int_{-\infty}^{\infty} x^2 \, g^2(x) \, dx \right).$$

(It comes up in a problem on PDE's I'm working on). I've written the same thing below using the $L_2$ norm:

$$ \left(|| xf || \cdot ||k \hat{g}||\right) + \left(||k\hat{f}|| \cdot || xg|| \right) $$

My initial guess was to use the Uncertainty Principle, but that only provides a lower bound for the product of $|| xf ||$ and $|| k \hat{f} ||$, which doesn't seem to help here. Are there any identities about the Fourier Transform that might help here? Perhaps, identities about $\int x^2 f^2 + \int k^2 \hat{f}^2$? Any help/suggestions would be greatly appreciated. Thanks!


For any $h\in L^2(\mathbb R)$, let $D(h)=\int_{\mathbb R} u^2|h(u)|^2du$.

You're looking for a lower bound for $D(f)D(\hat g)+D(\hat f )D(g)$. It's worth noting that, without any constraint on $f$ and $g$, the lower bound is $0$ since you can plug in $f=0$ or $g=0$. In fact the whole expression scales linearly with $\|f\|$ or with $\|g\|$. So let's assume that we've normalized $f$ and $g$ so that $\|f\|=\|g\|=1$.

By the Uncertainty Principle, there is a universal constant $C>0$ such that for all $h\in L^2(\mathbb R)$ with $\|h\|=1$, $$D(h)D(\hat h)\geq C$$

The exact value of $C$ depends on the normalization convention taken for the Fourier transform (it can be $\frac 1 {16\pi^2}$ or $\frac 1 {4\pi}$). The equality is attained for Gaussian functions.

With this, $$D(f)D(\hat g)+D(\hat f )D(g)\geq C\left(\frac{D(f)}{D(g)}+\frac{D(g)}{D(f)}\right)$$

The right-hand side is minimal iff $D(f)=D(g)$. In that case, the expression to minimize is equal to $2D(f)D(\hat f)$, which, by the Uncertainty Principle, is lower-bounded by $2C$, with equality attained by Gaussian functions.

We conclude that the lower bound is 2C, with equality attained when $f$ and $g$ are both Gaussian functions.

  • $\begingroup$ Wow! That's brilliant - thanks so much. Appreciate the help :) $\endgroup$ – chaad Sep 14 '20 at 2:16
  • $\begingroup$ You're too kind! $\endgroup$ – Stefan Lafon Sep 14 '20 at 2:17

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