Timeline for Concentration of measure and bounds on variance
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2011 at 15:10 | comment | added | Did | @t_t I see. Then controlling $E(\|\nabla F\|^2)$ might be an issue... | |
| Feb 9, 2011 at 12:44 | comment | added | Mikhail Kagalenko | Didier - this is quite correct, the gradient does go to infinity at $\mathbb{x}_0$. That's what makes the question interesting, because the <b>variation</b> remains bounded. It is qualitativerly clear why, but I would like to produce quantitative bound. | |
| Feb 9, 2011 at 6:30 | comment | added | Did | @t_t My point is that if $f(x_0)=0$ and $\|\nabla f(x)\|\cdot f(x)\to1$ when $x\to x_0$, then $\|\nabla f(x)\|\to+\infty$ when $x\to x_0$. Oh well. | |
| Feb 8, 2011 at 23:22 | comment | added | Mikhail Kagalenko | Didier - It's not a hypothesis, it's a property of the specific function that I am analyzing. | |
| Feb 8, 2011 at 22:57 | comment | added | Did | @t_t The hypothesis $\|\nabla f\| f\to1$ is even stranger. If you want to say that $f$ is $C^2$ and that $\nabla f(\mathbb{x}_0)=\mathbb{0}$, you could just say so. | |
| Feb 8, 2011 at 22:37 | vote | accept | Mikhail Kagalenko | ||
| Feb 8, 2011 at 22:30 | answer | added | Tom LaGatta | timeline score: 2 | |
| Feb 8, 2011 at 20:41 | history | edited | Mikhail Kagalenko | CC BY-SA 2.5 |
Fixed errors in the original question
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| Feb 8, 2011 at 20:33 | comment | added | Mikhail Kagalenko | The function is defined implicitly, so I mixed different gradients | |
| Feb 8, 2011 at 20:31 | history | edited | Mikhail Kagalenko | CC BY-SA 2.5 |
deleted 42 characters in body
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| Feb 8, 2011 at 20:26 | comment | added | Mikhail Kagalenko | Didier - I mean that $f(\mathbf{x})=(\nabla{}f)\cdot{(\mathbf{x}-\mathbf{x}_0)}+O((\mathbf{x}-\mathbf{x}_0)^2)$ near $\mathbf{x}_0$, but my notation was somewhat erroneous. Fixed that. | |
| Feb 8, 2011 at 18:14 | comment | added | Did | @tristes_tigres The hypothesis that $\|\nabla f(\mathbb{x})\|\to0$ and $f(\mathbb{x})\to0$ with $\|\nabla f(\mathbb{x})\|/f(\mathbb{x})\to1$ when $\mathbb{x}\to0$, is strange. Is this what you wanted to write? | |
| Feb 8, 2011 at 17:25 | history | asked | Mikhail Kagalenko | CC BY-SA 2.5 |