Nicolas Essis-Breton
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 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 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