Nicolas Essis-Breton
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 Oct 12 asked Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$ Oct 9 revised Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges error in dirichlet summation per davide comment Oct 9 accepted Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges Oct 9 asked Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges Oct 9 accepted Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. Oct 9 revised Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. 1/2 factor added per davide answer Oct 9 comment Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. @Sasha your comment mad me realize I didn't grasp the book argument. I added the complete argument plus the edits ou suggest. Thank you. Oct 9 revised Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. added more detail from the book, per sash comment Oct 9 asked Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. Sep 25 comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$ @did I believe you are pretty strong, finding this lower bound on the spot. Any idea for the original question? Sep 25 comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$ @did We showed $\sqrt{1+|R|^{-2}-2\Im(R^{-1}e^{-i\theta})} \ge 1-|R|^{-1}$. But I need a lower bound for $\sqrt{1+R^{-2}e^{-2i\theta}}$ to apply the DCT. What is the link between these? Sep 25 accepted Show translation is not continuous in $\text{Lip}_\alpha(T)$ Sep 24 comment Show translation is not continuous in $\text{Lip}_\alpha(T)$ @did Can you help me with the translation of continuity? I try to work $\frac{|f(t+h)-f(t+c+h)-f(t)+f(t+c)|}{|h|^\alpha}$, but I get nowhere. Sep 24 comment Show translation is not continuous in $\text{Lip}_\alpha(T)$ @did I don't see how you get $\|f\|=1$. I have $\|f\| \le \frac{\sqrt{|h|}}{|\sqrt{x+h}+\sqrt{|x|}|}$. Sep 24 revised Show translation is not continuous in $\text{Lip}_\alpha(T)$ state that T is the unit cirlcle Sep 24 comment Show translation is not continuous in $\text{Lip}_\alpha(T)$ @did I want $|x|$ to be the absolute value of x, for example $|-\pi|=\pi$. Then the map $x \mapsto \sqrt{|x|}$ is well-defined on $T$. Sep 24 comment Show translation is not continuous in $\text{Lip}_\alpha(T)$ @DavideGiraudo Yes, $T$ is the unit circle. Sep 24 asked Show translation is not continuous in $\text{Lip}_\alpha(T)$ Sep 23 comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$ @did I see your point. How do I get a lower bound for this radius? Sep 23 comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$ @did So $\|1+iR^{-1}e{-i\theta}\|^2=1+R^{-2}e^{-2i\theta} \ge 1$, and DCT follows. Is it correct?