Timeline for Is the tensor product of distributions a continuous bilinear map with respect to the weak topology?
Current License: CC BY-SA 4.0
9 events
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| May 20, 2021 at 18:05 | comment | added | Abdelmalek Abdesselam | Ah, good question. I retagged the post so hopefully experts on topological vector spaces can help. These notes seem to have an interesting discussion of the operation of double orthogonals asc.tuwien.ac.at/~enigsch/lehre/lcs.pdf | |
| May 20, 2021 at 18:02 | history | edited | Abdelmalek Abdesselam |
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| May 20, 2021 at 14:21 | comment | added | user449595 | I agree that the weak topology is the wrong one. The reason I asked the question is because I have a situation where I have sets of distributions $\mathcal{S}\subseteq\mathcal{D}'(X)$ and $\mathcal{T}\subseteq\mathcal{D}'(Y)$, and I want to show that the image of $(\mathcal{S}^\perp)^\perp\times (\mathcal{T}^\perp)^\perp$ in $\mathcal{D}'(X\times Y)$ is contained in $((\mathcal{S}\boxtimes\mathcal{T})^\perp)^\perp$. | |
| May 20, 2021 at 14:15 | vote | accept | user449595 | ||
| May 19, 2021 at 21:29 | answer | added | Abdelmalek Abdesselam | timeline score: 7 | |
| May 19, 2021 at 20:14 | comment | added | Abdelmalek Abdesselam | That's the point, it's not true. Why do you need the weak topology? It's the wrong one to use for this kind of question. | |
| May 19, 2021 at 19:19 | comment | added | user449595 | If $S_\alpha$ and $T_\beta$ converge weakly to $S$ and $T$ in $\mathcal{D}'(X)$ and $\mathcal{D}'(Y)$, respectively, then $S_\alpha\boxtimes T_\beta$ converges pointwise to $S\boxtimes T$ on the dense subspace of $\mathcal{D}(X\times Y)$ spanned by all functions of the form $f\boxtimes g$. But, if it is true, I do not see how to show that $S_\alpha\boxtimes T_\beta$ converges pointwise to $S\boxtimes T$ on all of $\mathcal{D}(X\times Y)$. | |
| May 19, 2021 at 16:48 | comment | added | Abdelmalek Abdesselam | If one uses the correct topology which is the strong dual topology then the map is continuous. I don't think it is continuous if one equips these spaces with the weak-$\ast$ topology. By continuous I mean continuous, not sequentially continuous. | |
| May 19, 2021 at 15:33 | history | asked | user449595 | CC BY-SA 4.0 |