1,672 reputation
1817
bio website nicolasessisbreton.com
location Montreal, Canada
age 30
visits member for 3 years, 1 month
seen 2 days ago

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


Aug
1
awarded  Yearling
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Our answers are complementary. But You were of precious help and I think you deserve the points. I learn a lot. Thank you.
Jun
28
accepted Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie I improved the first paragraph of the proof. You are of great help. Thank you for sharing your insight.
Jun
28
revised Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
First paragraph of the proof improved per Cameron comment
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie I think I can correct the flaw as follow. Assuming $g$ is discontinuous at $a$, there is a sequence $a_n$ in $A$ converging to $a$ and an $\epsilon >0$ such that for each $n$, $b_n=g(a_n)$ and $|b_n-g(a)|>\epsilon$.
Jun
28
answered Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
27
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie Thanks for your edit. You read my mind. I don't understand why, though, the $\arg_{b \in B}$ notation is not equivalent to the $B_a$ notation.
Jun
27
asked Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
23
revised Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
change X_t, Y_t to X,Y, added explanation of quasilinear, per Davide comment and did answer
Jun
23
comment Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
Strong as always. Thank you did. It's true that stochastic processes are not revelant. I got confuse in adapting my problem. I will correct the question.
Jun
23
accepted Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
Jun
22
comment Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
@DavideGiraudo I mean it's both quasiconcave and quasiconvex. Maybe only one of these is possible though.
Jun
22
asked Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
Jun
19
revised Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
1 char in body, following Byron comment, so that all $t$ are nonnegative
Jun
19
comment Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
@ByronSchmuland Good point. I forgot to mention that the $t$ are positive. Thank you.
Jun
19
comment Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
Both your answer and the one of copper.hat are excellent. I throw a coin to decide the winner.
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