Timeline for Is the following sequence convergent in the weak topology?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15, 2017 at 9:19 | comment | added | Anthony Quas | OK. Your answer seems exactly right. | |
| Feb 12, 2017 at 21:45 | comment | added | Nate Eldredge | @AnthonyQuas: In this topology, it doesn't converge to the zero measure; see above. (You might be thinking of the weak-* topology on $C_0(\mathbb{R})$.) | |
| Feb 12, 2017 at 20:54 | comment | added | Anthony Quas | I think this is convolution of a uniform probability measure on $[0,\tau_n]$ with $\mu$; this sequence isn't tight and converges to the 0 measure. | |
| Feb 12, 2017 at 16:25 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
added 218 characters in body
|
| Feb 12, 2017 at 16:06 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
added 670 characters in body
|
| Feb 12, 2017 at 15:47 | comment | added | Nate Eldredge | @AnthonyQuas: Oh, I didn't read the question carefully and I was thinking of $\tau_n \to 0$. But note that $f$ is not assumed to be compactly supported. Let me look at this again when I have more time. | |
| Feb 12, 2017 at 9:08 | comment | added | Anthony Quas | Hold on a minute! Surely the convergence is just to 0. If $f$ is any compactly supported function, then $|\int f(x+t)\,d\mu|<\epsilon$ for all $t$ outside a bounded range. | |
| Feb 11, 2017 at 18:52 | vote | accept | g.pomegranate | ||
| Feb 11, 2017 at 18:16 | history | answered | Nate Eldredge | CC BY-SA 3.0 |