Timeline for Uncertainty principle: minimize $\int_{-\infty}^\infty |t| |\widehat{f}(t)|^2 dt$ for $f$ of compact support
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| yesterday | comment | added | H A Helfgott | @ChristianRemling What you are saying is a qualitative restatement of the problem (of course we need to concentrate $\widehat{f}$ near $0$, obeying the constraints). At any rate, it seems to me that one avenue of attack is to discretize the original problem. Let me do that now. (Do you have another approach?) | |
| yesterday | comment | added | H A Helfgott | My one remaining misgiving lies in the reduction of minimizing $\int_{-\infty}^\infty |t| |\hat{f}(t)|^2 dt$ to minimizing $\int_{-\infty}^\infty |t|^2 |\hat{f}(t)|^2 dt$. This wasn't really rigorous, as the space of functions we are considering was not closed under $B$. (Incidentally, that is also true for $f\mapsto f'$; notice that our solution to that problem (which does seem sound) is not an eigenfunction: $\cos'$ is not a multiple of $\cos$, though $\cos$ feels 'eigenfunctiony'.) | |
| yesterday | comment | added | H A Helfgott | Ah, the problem of minimizing $|f'|^2$ has a nice solution! Consider first a discretization - let us just consider $f$'s that are piecewise linear - more specifically, linear on all intervals [a/N-1/2,(a+1)/N-1/2], and 0 at -1/2 and 1/2. Then it's not hard to reduce the problem of minimizing $|f'|^2$ to that of finding the top eigenvalue of the adjacency matrix on a path graph (h/t @WillSawin for seeing this). The top eigenfunction is indeed equal to cos(x) at all a/N-1/2, 0<=a<=N. The limit is obviously $\cos(x)$. PS. One can show one can always discretize by $|f'|_{TV}<\infty$. | |
| yesterday | comment | added | H A Helfgott | @juan I meant $\hat{f}(t) = O(1/t^2)$. | |
| yesterday | comment | added | juan | I am confused you speak of support of $f$ contained in $[-1/2,1/2]$ and $f=O(1/t^2)$. | |
| yesterday | comment | added | H A Helfgott | ... though $f'\ast \frac{1}{x}$ blows up at both $-1/2a$ and $1/2a$ for $f(x) = \cos(ax) 1_{[-1/2a,1/2a]}$, meaning we have to tread carefully. | |
| yesterday | comment | added | H A Helfgott | However, for $\cos(a x)\cdot 1_{[-1/2a,1/2a]}$, $|f'|_2^2$ is still meaningful, and in the sense of distributions or measures, so is $\langle f,f''\rangle$. | |
| yesterday | comment | added | H A Helfgott | Here I want to conclude by saying 'the eigenfunctions of $f\mapsto f'$ are the exponentials' but of course they are not of compact support. The issue is that $e^{i a x}\cdot 1_{[-b,b]}$ is almost an eigenfunction of $f\mapsto f'$, except it's not continuous at $b$ or $-b$, and $\cos(a x)\cdot 1_{[-1/2a,1/2a]}$ is continuous and almost an eigenfunction of $f\mapsto f''$, except it is not differentiable at $-1/a$ or $1/a$. | |
| yesterday | comment | added | H A Helfgott | Here is a possible approach. Let $B$ the linear operator such that $\widehat{Bf}(t) = |t| \widehat{f}(t)$. (Namely: $B$ is given by $Bf = f'\ast \frac{1}{2\pi^2 x}$.) We are being asked to minimize $\langle Bf, f\rangle$ given $|f|_2=1$, i.e., find the lowest eigenvalue. Since the eigenvalues are clearly all positive, this is the same as minimizing $|B f|_2$. However, $$|B f|_2 = \int_{-\infty}^\infty ||t| \widehat{f}(t)|^2 dt = \int_{-\infty}^\infty |t \widehat{f}(t)|^2 dt,$$ which equals both $\frac{1}{(2\pi)^2} |f'|_2^2$ and $- \frac{1}{(2\pi)^2} \langle f,f''\rangle$. (continued below) | |
| yesterday | history | asked | H A Helfgott | CC BY-SA 4.0 |