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| visits | member for | 1 year, 7 months |
| seen | 8 mins ago | |
| stats | profile views | 483 |
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1h |
accepted | Continuity of $d(x,A)$ |
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2h |
revised |
Continuity of $d(x,A)$ added 146 characters in body |
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2h |
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Continuity of $d(x,A)$ @Martin Yes thank you for pointing that out. If $X=\mathbb{R}^n$ and $A$ is closed then what I said is true. (Proof uses compactness of unit ball). |
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2h |
asked | Continuity of $d(x,A)$ |
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May 20 |
comment |
What's the exact definition for convolution? @vanabel I have thought about it for a while and haven't come up with a proof or counterexample. The standard result on existence of convolutions (on $\mathbb{R}^d$) is called Young's inequality: If $f \in L^p(\mathbb{R}^d), g \in L^q(\mathbb{R}^d)$ and $\frac1p + \frac1q = \frac1r + 1$ then $||f * g||_r \leq ||f||_p||g||_q$. |
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May 19 |
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What's the exact definition for convolution? The convolution is NOT well defined for all $f,g$ measurable. Choose $f,g$ such that $f(x-y)g(y)$ is not integrable for any $x$. E.g. $f(x)=1, g(x)=\sin(x)$ then clearly the convolution $f*g$ doesn't make sense for any $x$. |
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May 17 |
awarded | Nice Question |
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May 17 |
asked | Examples of properties that hold almost everywhere, but that explicit examples unknown. |
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May 17 |
answered | A basic question on the definition of $E[X]$ |
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May 14 |
comment |
Extending the multiplication property of the Fourier transform to $L^2(\mathbb{R})$ The usual method I have seen is to show the result holds for all Schwarz functions and then use approximate. |
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May 10 |
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A few finance probability and logic questions. Help? I did some work, but I'm stuck on a few. From my experience, a large percentage of the questions I have been asked for quant positions came directly out of this book (it is listed on the website you linked as well). amazon.com/gp/product/0486653552?tag=quantfinaneng-20 |
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May 10 |
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If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ Then from here, I am using that if $\lim a_n$ exists, then $\lim a_n = \liminf a_n$, and if $a_n \leq b_n$ for all $n$ , then $\liminf a_n \leq \liminf b_n$. These are basic facts from real analysis which you should be very familiar with before attempting the more subtle theory involved in functional analysis. |
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May 10 |
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Probability of $X^{2} > Y^{3}$ over distribution other than uniform @Katarzyna Yes it sounds like you have the general idea correct. However, it is clear that you have not yet learned how to formulate probability in a rigorous setting. There is nothing wrong with this for now, but without the tools of measure theory at your disposal, it is easy to ask a simple question (like this one) which has a simple answer, but that understanding of why the answer is correct is beyond the scope of your knowledge. If you are really interested in understanding probability, then you should definitely look into taking measure-theoretic probability. |
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May 10 |
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If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ @Fractal20 The statement $x_n \to x$ weakly by definition means $x^*(x_n) \to x^*(x)$ for every $x^* \in X^*$. The absolute value function $|\cdot|$ is continuous on $\mathbb{R}$ so we also have $|x^*(x_n)| \to |x^*(x)|$. Thus the first limit is defined. |
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May 9 |
revised |
If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ added 120 characters in body |
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May 9 |
comment |
If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ @Fractal20 For your second comment, yes $||x^*|| = \sup \{ x^*(x), x \in X, ||x||\leq 1\}$ by definition. However, it follows quickly from the Hahn-Banach theorem that for any $x \in X$ there is an $x^* \in X^*$ such that $x^*(x) = ||x||$. Thus it follows (convince yourself of this) that for any $x \in X$ we have $||x|| = \sup\{x^*(x) : x^* \in X^*, ||x^*|| \leq 1\}$. |
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May 9 |
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If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ @Fractal20 As for your first comment, we know $||x^*(x_n)|| \leq ||x^*||\ ||x_n||$ for all $n$. However, we do not know that $||x_n||$ converges. Thus, the best we can say is that $\lim |x^*(x_n)| = \liminf |x^*(x_n)| \leq \liminf ||x^*||\ ||x_n||$. |
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May 8 |
revised |
If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ added 88 characters in body; edited tags |
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May 8 |
revised |
If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ added 174 characters in body |
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May 8 |
answered | If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$ |
