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 Apr16 awarded Nice Question Dec3 awarded Popular Question Sep24 awarded Autobiographer Jul2 awarded Curious Apr23 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. Apr18 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. ：） Apr18 asked Linear Fractional Transforms maps the upper half unit disc onto the first quadrant Apr17 accepted Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$? Apr15 answered Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$? Apr15 awarded Informed Apr14 asked Continuous piecewise smooth function $=$ a globally $\mathcal{C}^1$ function $+\sum a_i|s-\alpha_i|$? Apr14 accepted An inequality of J. Necas Nov10 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 8 characters in body Nov10 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 $p$. Nov10 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? Nov10 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 Nov10 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. Nov10 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? Nov8 awarded Yearling Oct17 accepted About the trace of Sobolev functions