Timeline for On effective constructions in the functional analysis of Volterra's integration operator
Current License: CC BY-SA 4.0
16 events
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| Jul 19, 2018 at 16:03 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 18, 2018 at 8:31 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 18, 2018 at 8:21 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 18, 2018 at 7:34 | comment | added | Vesselin Dimitrov | @fedja: Now I see - this is neat - we may simply extract a limit $Vf_n \to Vf$ from a sequence of hypothetical counterexamples, make a construction (2) for $f$ and just take an $f_n$ farther enough along the sequence to derive the contradiction. I was a little confused at first as I presumed the argument would start with choosing some finite $\epsilon$-net from the compact set $\{ Vf \mid |f| \leq L \}$, whereas we are really letting $\epsilon \to 0$ and passing to a sequence for the contradiction, even with the fixed value of $\varepsilon$. | |
| Jul 18, 2018 at 1:47 | comment | added | fedja | It is used not for compactness but for the statement that every function in the closure of the corresponding family of $Vf$ generates the entire space while in the first counterexample $0$ was in the closure (which is exactly what I used). | |
| Jul 17, 2018 at 22:45 | comment | added | Vesselin Dimitrov | Yes, the previous example demonstrated that it certainly makes a difference that the approximation is intended in $L^2$ and not in $C$. Thanks very much, this appears to be making sense, the one thing that bothers me on a first look is how the lower bound of the condition $1 \leq f \leq L$ (clearly necessary for the $C(L,\varepsilon)$ statement to hold) gets used in the compactness argument with $\{ Vf \mid 1 \leq f \leq L \}$. | |
| Jul 17, 2018 at 21:50 | comment | added | fedja | You are right. For some reason I thought that we were approximating in $C$ rather than in $L^2$. Then some $C(L,\varepsilon)$ exists. Indeed, after the first application of $V$ and passing to the closure, you get a compact family of functions each of which generates the entire space (in $L^2$, but not in $C$). So just find some coefficients for each of them and use the fact that they work in some neighborhood. | |
| Jul 17, 2018 at 19:29 | comment | added | Vesselin Dimitrov | (We just abstain from using $V^0(f) = f$ in this example, but make use of the cancellations from the oscillations of $f$ in the further iterates $V^1(f), V^2(f), \ldots$. I agree that $a_0 \approx 0$ necessarily in this example.) | |
| Jul 17, 2018 at 16:54 | comment | added | Vesselin Dimitrov | @fedja: I am confused by this example, for aren't the small strictly positive ($k \neq 0$) iterates $V(\cos{Nx}) = N^{-1} \sin(Nx), \, V^2(\cos(Nx)) = N^{-2} - N^{-2}\cos(Nx)$ etc., all essentially negligible in magnitude (while of course wildly oscillating - that doesn't harm when the magnitude is this small), once $N$ is very large? We are then reduced to approximating the constant $1$ in $L^2(0,1)$ by a small degree polynomial without free term and with fairly small coefficients, which is possible. I apologize if I am overlooking something again. | |
| Jul 17, 2018 at 16:07 | comment | added | fedja | Still bad. Take $f(x)=2-\cos Nx$. For the same reason, as before, under condition (1), large $k$ do not matter on $[0,0.5]$ while small $k\ge 1$ give you almost a polynomial, which, since it has bounded degree, cannot catch up with the oscillations of $\cos$ for large enough $N$... | |
| Jul 16, 2018 at 8:32 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 16, 2018 at 8:31 | comment | added | Vesselin Dimitrov | @fedja: Thank you for pointing this out: the condition $f(0) = 1$ and $|f| \leq L$ is indeed not the relevant one, in view of these oscillating examples. I was thinking that I should bound $f$ from below somehow, not only at $0$. For my example I can assume $1 \leq f \leq L$ on all of $[0,1]$; I'll edit as this amendment seems to make a difference. In any case, I am curious about any way of quantifying (2), in terms of some such relevant condition about $f$. | |
| Jul 15, 2018 at 22:47 | comment | added | fedja | That would be a bit too much because the norm of $V^k$ in $C[0,0.5]$ is only $\frac 1{k!}2^{-k}$ and you can kill as many low powers as you wish by taking a fast oscillating function (like $\cos Nx$ with huge $N$), so if this is what you really need, you are out of luck. If you can get away with less, let us know what it is. | |
| Jul 15, 2018 at 17:31 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 15, 2018 at 17:21 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Jul 15, 2018 at 17:10 | history | asked | Vesselin Dimitrov | CC BY-SA 4.0 |