Timeline for Convergence of some object depending on functions with compact support
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2018 at 10:40 | comment | added | YCor | Yes. Actually you're using right-invariance of $\mu$ and not left as I did. In this case the right-regular representation is unitary. Said otherwise: for $G$ possibly non-unimodular and $\mu$ left-invariant, my argument works for the right-regular representation, and your argument works for the left-regular representation (these two representations are of distinct interest when $G$ is non-unimodular; only the left one is unitary). | |
| Mar 31, 2018 at 10:09 | comment | added | Constantin K | Thank you very much. To simplify a little: $$\begin{align*} ||\pi(\varphi)f||_2^2 &= \int_G |(\pi(\varphi)f)(x)|^2 \, d\mu(x) \\ &= \int_G \bigg| \int_{\mathrm{supp}(\varphi)} \varphi(g)f(xg) \, d\mu(g) \bigg|^2 \, d\mu(x) \\ & \leq ||\varphi||_2^2 \int_G\int_{\mathrm{supp}(\varphi)} |f(xg)|^2 \, d\mu(g)d\mu(x) \\ &= ||\varphi||_2^2 \int_{\mathrm{supp}(\varphi)} \int_G |f(xg)|^2 \, d\mu(x)d\mu(g) \\ &= ||\varphi||_2^2 \int_{\mathrm{supp}(\varphi)} \int_G |f(x)|^2 \, d\mu(x)d\mu(g) \\ &= \mu(\mathrm{supp}(\varphi)) \cdot ||\varphi||_2^2 \cdot ||f||_2^2 < \infty. \end{align*} $$ | |
| Mar 31, 2018 at 8:54 | vote | accept | Constantin K | ||
| Mar 31, 2018 at 8:33 | history | answered | YCor | CC BY-SA 3.0 |