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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?
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 This is the radius of the complex number $1+iR^{-1}e^{-i\theta}$.
Sep
23
asked Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
Sep
16
accepted Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
revised Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
corrected question per Alex Becker comment
Sep
16
comment Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
@AlexBecker Good eyes, I meant $\text{Im}(f(t))$ is identically zero. Thank you.
Sep
16
asked Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
accepted Does the complex conjugate of an integral equal the integral of the conjugate?
Sep
16
asked Does the complex conjugate of an integral equal the integral of the conjugate?