# Y.Z

Shanghai, China

Age: 25

First year graduate student in PDE.

 Apr 16 awarded Nice Question Dec 3 awarded Popular Question Sep 24 awarded Autobiographer 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?added 8 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: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