1,648 reputation
616
bio website nicolasessisbreton.com
location Montreal, Canada
age 29
visits member for 2 years, 8 months
seen 22 hours ago

I'm a graduate math student at Concordia University, Montreal.


Jun
19
accepted Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
Jun
19
comment Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
@copper.hat I meant the integral of $x^t$, sorry.
Jun
19
revised Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
2 char in title and body, following Nimza comments
Jun
19
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.
Jun
19
asked Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
Jun
15
accepted Finding the fixed point of a function
Jun
14
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?
Jun
14
comment Finding the fixed point of a function
which theorem do you have in mind?
Jun
14
asked Finding the fixed point of a function
Apr
12
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 $|g|\le 1$
Apr
11
awarded  Teacher
Apr
11
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.
Apr
11
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 $|g|\le 1$
Apr
10
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 $\|f\|_p$.
Apr
10
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 $c>0$.
Apr
10
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}
Apr
10
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 $\mathbb R \setminus E$.
Apr
10
asked Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
Apr
4
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 $g=f\chi_{\mathbb{R} \setminus E}$.
Apr
4
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 $|g|\le 1$