Timeline for Can convolution on $R_+$ be discontinuous everywhere ?
Current License: CC BY-SA 3.0
2 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2012 at 0:59 | comment | added | Ollie | Certainly $f*g$ is locally $L^1$, so almost all points are also Lebesgue points for $f*g$. For every such point $x$ and every $\varepsilon>0$ there exists a $\delta>0$ with almost all points $y\in B(x,\delta)$ satisfying $|f*g(x)-f*g(y)|<\varepsilon$. (Why? Just assume not...) So there exists a version of $f*g$ continuous at $x$. With a little thought you can now devise a scheme to modify $f*g$ so it really is continuous except at countably many points. | |
| Mar 31, 2012 at 21:45 | history | answered | Gerald Edgar | CC BY-SA 3.0 |