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 Yearling
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Jul
17
awarded  Yearling
Apr
16
awarded  Nice Question
Dec
3
awarded  Popular Question
Sep
24
awarded  Autobiographer
Jul
2
awarded  Curious
Apr
23
comment Linear Fractional Transforms maps the upper half unit disc onto the first quadrant
@LeeMosher: I feel ashamed for asking this question. I should read my textbook more carefully. The point here is that LFT preserve angles on the whole complex plane except one possible point. The upper half disc has two right angles on the boundary, then one of them must be map to the origin. Say, $z=-1$, then $z=1$ must be map to $\infty$, and then the LFT is not analytic at $z=1$. So, there is no such an "angle preserving" at $z=1$. And hence my question does not make sense. I hope I get the point.
Apr
18
comment Linear Fractional Transforms maps the upper half unit disc onto the first quadrant
@LeeMosher:I think there are two such LFTs, $T(z)=k(1+z)/(1-z),k>0$ and $T(z)=ih(1-z)/(1+z),h>0$. If you wanna me choose one, I wanna say: Both. :)
Apr
18
asked Linear Fractional Transforms maps the upper half unit disc onto the first quadrant
Apr
17
accepted Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$?
Apr
15
answered Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$?
Apr
15
awarded  Informed
Apr
14
asked Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$?
Apr
14
accepted An inequality of J. Necas
Nov
10
revised Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
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Nov
10
comment Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
@DanielFischer:You're right!In fact, I just need the case when $p=3$. I've changed the assumption of $p$.
Nov
10
comment Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
@DanielFischer:But what if $u,v$ change signs?
Nov
10
revised Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
added 55 characters in body
Nov
10
comment Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
@DanielFischer: Yes,thanks.
Nov
10
asked Does $\left|\left(\int_{\Omega}u^p\right)^{1/p}-\left(\int_{\Omega}v^p\right)^{1/p}\right|\leq C\left(\int_{\Omega}|u-v|^p\right)^{1/p}$ hold?
Nov
8
awarded  Yearling