Nicolas Essis-Breton
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 Jun19 comment Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure @Nimza Good eyes, I should have wrote $x^t$, not $x$. Thank you. Jun19 asked Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure Jun15 accepted Finding the fixed point of a function Jun14 comment Finding the fixed point of a function I'm not clear on how to interpret mixed strategies in my case. The strategy sets $A$ and $B$ are both convex subset of $\mathbb{R}^n$. Let $\Pi(A)$ be the set of possible probability distribution on $A$. My understanding is that the mixed strategies are $\Pi(A) \times \Pi(B)$. Aren't these different? Jun14 comment Finding the fixed point of a function which theorem do you have in mind? Jun14 asked Finding the fixed point of a function Apr12 accepted Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$ Apr11 awarded Teacher Apr11 comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$ @leo Good idea. It will enhance the post. I did it. Hope it's correct. Apr11 answered Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$ Apr10 comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$ @DavideGiraudo I think any kind of inequality will do as long as it involves $\|f \chi_E \|_r$, $\|f \chi_{\tilde E}\|_s$ and $\|f\|_p$. Apr10 comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$ @DavideGiraudo This point is not clear from the question. With the work I did, I think the inequality should be of the form $\|f \chi_E\|_r+ \|f\chi_{\tilde E}\|_s \le c \|f\|_p$ for some $c>0$. Apr10 revised Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$ 9 characters: added meaning of {\tilde E} Apr10 comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$ @MattN. Sorry, it's $\mathbb R \setminus E$. Apr10 asked Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$ Apr4 comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$ @lazyhaze I see, thanks. It suffices to take $g=f\chi_{\mathbb{R} \setminus E}$. Apr4 asked Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$ Mar9 comment Brownian motion: hitting time of double barrier vs hitting time of single barrier Our textbook refers to these as the heat equations (Introduction to Stochastic Process, Lawler). The precision you add on this in your post, helps me better understand these equations. Thank you very much Didier. Mar9 comment Brownian motion: hitting time of double barrier vs hitting time of single barrier how do you get $\partial_t u=\frac12 \partial_{zz}^2u$? I know the reflexion principle, and the Markov property of a Brownian motion, but neither seem to justify this statement. Mar8 comment Brownian motion: hitting time of double barrier vs hitting time of single barrier For the continuity of $z\mapsto\mathrm P_z(T\gt1)$. This seems intuitive as a Brownian motion has continuous path, but without deriving the distribution of $P_z(T\gt1)$, can it be prooved?