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In harmonic analysis, we have the Heisenberg uncertainty principle, i.e.

Suppose $f\in\mathcal{S}(\mathbb{R})$, which satisfies the normalizing condition $\int_{\mathbb{R}}|f(x)|^2dx=1$, then $$\left(\int_{\mathbb{R}}x^2 |f(x)|^2 dx \right)\left(\int_{\mathbb{R}}\xi^2 |\hat{f}(\xi)|^2 d\xi \right)\geq \frac{1}{(4\pi) ^2}.$$ Where $\hat f(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x \xi}dx.$

And this formula can be generalized to $n$-dimension:

Suppose $f\in\mathcal{S}(\mathbb{R}^n)$, which satisfies the normalizing condition $\int_{\mathbb{R}^n}|f(x)|^2dx=1$, then $$\left(\int_{\mathbb{R}^n}|x|^2 |f(x)|^2 dx \right)\left(\int_{\mathbb{R}^n}|\xi|^2 |\hat{f}(\xi)|^2 d\xi \right)\geq \frac{n^2}{(4\pi) ^2}.$$ Where $\hat f(\xi)=\int_{\mathbb{R}^n}f(x)e^{-2\pi i x\cdot\xi}dx.$

Since we still have $$\left(\int_{\mathbb{R}^n}x_i^2 |f(x)|^2 dx \right)\left(\int_{\mathbb{R}^n}\xi_i^2 |\hat{f}(x)|^2 dx \right)\geq \frac{1}{(4\pi) ^2}.$$ Then, we derive the generalized inequality by using the Cauchy-Schwartz inequality.

And the equalities can be achieved: Consider the Gaussian distribution function and adjust the coefficients suitably.

Now, I am looking for a sequence of functions $\{f_k\}_{k\geq 1}\subset \mathcal{S}(\mathbb{R}^n)$, each $f_{k}$ satisfies the normalizing condition, but for $i\neq j$, we have $$\left(\int_{\mathbb{R}^n}x_i^2 |f(x)|^2 dx \right)\left(\int_{\mathbb{R}^n}\xi_j^2 |\hat{f}(x)|^2 dx \right)\to 0, \quad (k\to \infty).$$ Maybe we can construct such sequence by adjusting some coefficients of the Gaussian distribution function? Any help would be appreciated.

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  • For completeness, the integral inequality you state is proven here in one dimension and here in multiple dimensions. (In fact, the proof in multiple dimensions that I cite is sloppy because it doesn't state in which function space the identities used hold and where they are inferred by continuity arguments, but oh well...) – Maximilian Janisch Jan 04 '22 at 14:47

1 Answers1

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We consider the function with the form $$ f=c\cdot e^{-(k_1 x_1^2+k_2 x_2^2+x_3^2+\cdots+x_n^2)/2} $$ and then adjust the coefficients. First, since $$ \Vert f\Vert_{L^2}=1$$ we have \begin{align*} 1 &=c^2\cdot \Big(\int e^{-k_1 x_1^2}\, dx_1\Big)\Big(\int e^{-k_2 x_2^2}\, dx_2\Big)\Big(\int e^{-x_3^2}\, dx_3\Big)\cdots \Big(\int e^{- x_n^2}\, dx_n\Big) \\ &= c^2\cdot \big(\frac{\pi}{k_1}\big)^{1/2}\cdot \big(\frac{\pi}{k_1}\big)^{1/2}\cdot \pi^{(n-2)/2} \end{align*} thus, we can take $$c=\Big(\frac{k_1 k_2}{\pi^n}\Big)^{1/4}. $$ Furthermore, we have \begin{align*} \int x_1^2 |f|^2\,dx &=c^2\Big(\int x_1^2 e^{-k_1 x_1^2}\,dx_1\Big)\cdot \Big(\int e^{-k_2 x_2^2}\,dx_1\Big)\Big(\int e^{-x_3^2}\, dx_3\Big)\cdots \Big(\int e^{- x_n^2}\, dx_n\Big)\\ &=\Big(\frac{k_1 k_2}{\pi^n}\Big)^{1/2}\cdot \frac{\pi^{1/2}}{2k_1^{3/2}}\cdot \frac{\pi^{1/2}}{k_2^{1/2}}\cdot \pi^{(n-2)/2}\\ &=\frac{1}{2k_1}, \end{align*} and \begin{align*} \int \xi_2^2|\hat{f}|^2\,dx &=\int |\partial_{x_2}f|^2\,dx\\ &=\int |k_2x_2\cdot f|^2\,dx\\ &=k_2^2\cdot \frac{1}{2k_2}\\ &=\frac{k_2}{2} \end{align*} thus $$\Big(\int x_1^2|f|^2\,dx\Big)\Big(\int \xi_2^2|\hat{f}|^2\,dx\Big)=\frac{k_2}{4k_1} $$ so, what we need is to take suitable coefficients $k_1,k_2$ such that $$\frac{k_2}{k_1}\to 0.$$

user1000039
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