Timeline for Eigenvalue of a convolution and a restriction?
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22 events
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| Dec 12 at 8:22 | comment | added | H A Helfgott | Asked the question at mathoverflow.net/questions/483993/… . Ah, maybe I can see an approach now, using precisely $\langle f,Bf\rangle = \langle Af,Af\rangle$. | |
| Dec 12 at 7:57 | comment | added | H A Helfgott | It's also obviously the case that, by asking which function $f$ with support on $[-1/2,1/2]$ (and $|f|_2=1$, $f$ absolutely continuous, etc.) minimizes $\int |t| |\widehat{f}(t)|^2 dt$, we are asking a question of uncertainty-principle type. | |
| Dec 12 at 7:36 | comment | added | H A Helfgott | Obvious variant: consider the operator $B:f\mapsto \frac{1}{2\pi^2} \left(f'\ast \frac{1}{x}\right)$. Then the Fourier transform of $Bf$ is $|t| \widehat{f}(t)$. Hence $\int_{-\infty}^\infty |t| |\widehat{f}(t)|^2 dt = \langle \widehat{f}, \widehat{B f}(t)\rangle = \langle f,B f\rangle$, and $(Bf)(x) = \frac{1}{2\pi^2} \int_{-\infty}^\infty \frac{f'(x+y)}{-y} dy$, which, for $f$ everywhere differentiable, equals $\frac{1}{2\pi^2} \int_{-\infty}^\infty \frac{f(y)-f(x)}{(y-x)^2} dy$. | |
| Dec 12 at 7:15 | comment | added | H A Helfgott | In the definition of $A$, $\frac{1}{\sqrt{x}}$ should be $\frac{1}{\sqrt{|x|}$. | |
| Dec 12 at 7:01 | comment | added | H A Helfgott | @LiorSilberman Yes, exactly: either you restrict in the definition of your operator, and then your operator is compact and the theory is beautiful - but we do not know what the eigenfunctions are (beyond knowing that they are not anything we've discussed!), or you do not restrict in the definition of your operator, and then you have access to eigenfunctions cos(ax) (beautiful but not L^2) - and your operator is not compact, and the theory is complicated (well, to us plebs). | |
| Dec 12 at 3:00 | comment | added | Lior Silberman | I see you've changed variables, so divide all my intervals by $2\pi$. | |
| Dec 12 at 2:58 | comment | added | Lior Silberman | Let C be the operator of convolution with $\eta$ (which we may take supported on $[-2\pi,2\pi]$). Then $C\colon L^2(-\pi,\pi)\to L^2(-3\pi,3\pi)$ is compact (continuous kernel, compact intervals). Let $R\colon L^2(-3\pi,3\pi)\to L^2(-\pi,\pi)$ be the restriction, a bounded operator. Then $A=RC$ is compact, and $A^*A = RC^2R$ is compact and self-adjoint, hence has discrete spectrum. | |
| Dec 11 at 23:39 | comment | added | H A Helfgott | But what are the eigenfunctions of $A$? Again we have the problem that, while there presumably are eigenfunctions supported in $1_{[-1/2,1/2]}$, it is easy to show that they can't be $0$ at the boundary. We can decide to look at eigenfunctions without the support condition. Then, for $a\geq 0$, $x\mapsto\cos(a x)$ is an eigenfunction of $A$ (with eigenfunction $i \sqrt{\frac{|a|}{2\pi}}$) but it's obviously in $L^2$. I guess we are not in the Kansas of compact operator theory any more. Are there other eigenfunctions? (Is there a discrete spectrum?) | |
| Dec 11 at 23:27 | comment | added | H A Helfgott | Ah! I now see how to do it (up to a $O(\epsilon^2)$ term). Consider the operator $A:f\mapsto \frac{1}{2\pi i} (f'\ast \frac{1}{\sqrt{x}})$. The Fourier transform of $A f$ is $\sqrt{|t|} \widehat{f}(t)$. Hence, $\int_{-\infty}^\infty |t| |\widehat{f}|^2 dt = \int_{-\infty}^\infty |\widehat{A f}(t)|^2 dt = \int_{-\infty}^\infty |(A f)(x)|^2 dx.$ Show by integration by parts that $(A f)(x) = \frac{1}{4\pi i} \int_{-\infty}^\infty \frac{f(x+y)-f(x)}{|y|^{3/2}} dy$. By either calculus of variations or compact-operator-something, $|A f|_2$ is minimized when $f$ is an eigenfunction of $A$. (cont.) | |
| Dec 11 at 22:18 | comment | added | H A Helfgott | @WillSawin I'm not sure I can see how this is possible. Wouldn't the discontinuity at $1$ (by a constant independent of $\epsilon$) cause the decay of $\widehat{f}$ to be only like $O(1/t)$, and $I_\epsilon(f)$ not be $O(\epsilon)$, just as above? | |
| Dec 11 at 21:55 | comment | added | Will Sawin | Actually maybe $\cos ( (\pi-\delta) x))$ beats $\cos(\pi x)$ for some $\delta>0$ independent of $\epsilon$. | |
| Dec 11 at 21:25 | comment | added | Will Sawin | Convolution with $\eta$. If $\eta * f$, restricted to the interval $[-1/2,1/2]$, is equal to $\lambda f$, and we shrink the support of $\eta$, which is like growing the interval, then maybe we can say that $\eta * f$ is approximately $\lambda f$ so $f$ is approximately an eigenfunction of convolution with $\eta$. | |
| Dec 11 at 21:12 | comment | added | H A Helfgott | @WillSawin What do you mean by an eigenfunction of the convolution operator? Convolution with what? | |
| Dec 11 at 20:27 | comment | added | Will Sawin | Roughly that away from the boundary your operator is well-approximated by just convolution, so we need an eigenfunction of the convolution operator, which would be a sine or cosine of something. We obtain a loss compared to the best possible $\lambda=1$ if it's an eigenfunction with value less than $1$, and a loss also at the boundary from the part of $f$ that after convolution lies outside $[-1/2,1/2]$ and is ignored. Lowering the frequency of the cosine lowers the first loss but raises the second, but only at a worthwile rate if you lower it a small amount. | |
| Dec 11 at 19:35 | comment | added | H A Helfgott | @WillSawin Probably a better reason to suspect $\cos$ is that it is an eigenfunction of $f\mapsto f''$ (except at $-1/2$ and $1/2$), and that these operators are starting to look awfully like $f\mapsto f''$ up to a first order of approximation. | |
| Dec 11 at 19:00 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Dec 11 at 18:45 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Dec 11 at 17:56 | comment | added | H A Helfgott | @WillSawin I was guessing basically as a slightly more advanced AI would: a truncated cosine is one of those things that actually appear as eigenfunctions, as I vaguely remember from many years of half-listening to talks in other areas. I don't remember anything about path graphs specifically - I'll have to learn. But yes, I was thinking that this is like an adjacency matrix, and a possible approach is simply to take a high power of it, i.e., iterate the operator many times on a fairly arbitrary function. But why do you suspect this might be optimal? | |
| Dec 11 at 16:27 | comment | added | Will Sawin | Your operator is somewhat similar to the adjacency matrix of a path graph, whose top eigenvector is a cosine function.Was this the source of cosine idea? Considering that example I think positivity should be obtained by just stretching out, i.e. $\cos ( (\pi - C \epsilon) x$ for appropriate $C$. I sort of suspect this example will give the optimal asymptotic for $(1-\lambda)/\epsilon$, though not the optimal exact value. (Although maybe stretching out does not affect the asymptotic). | |
| Dec 11 at 15:10 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Dec 11 at 14:50 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Dec 11 at 14:33 | history | answered | H A Helfgott | CC BY-SA 4.0 |