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 Aug12 asked Evolutionary algorithm Jul2 awarded Curious Jun28 accepted Exercise on isometry Jun20 comment Change of variable for Lebesegue Integral I don't know if this is right but for the theorem of change of variables between measures I have that I can consider a Borel measure $\mu (E)=\int_E G' d\lambda$ for each $E \in \mathbb{B}([a,b])$. Then for each $E' \in \mathbb{B}([c,d])$ we have $\mu (G^{-1}(E'))=\lambda (E')$. Now $\lambda(G^{-1}(N) \cap H) \leq \lambda(G^{-1}(N))=\lambda(N)=0$. Jun20 asked Change of variable for Lebesegue Integral Jun19 accepted Theorems on continuous embedding Jun19 comment Theorems on continuous embedding My definition is: $X \hookrightarrow Y$ if $X\subseteq Y$ and the map $J:X \rightarrow Y$ defined as $Jx=x$ is continuous: $||Jx||_Y=||x||_Y \leq C ||x||_X \>\>\> \forall x\in X$ with $C$ a nonnegative costant. $J$ assumed to be linear. Jun19 comment Theorems on continuous embedding Well I'm trying to use the definition of weak convergence and continuous embedding but I'm not able to reach the thesis. The first point seems to me naturally coming because if $A$ equals its closure in Y and $A\subset X \subset Y$ then $A$ equals its closure in X also. But I don't think this is rigorous and I'm not using the hypothesis of continuous embedding... Jun19 asked Theorems on continuous embedding Jun18 comment Riesz Lemma for reflexive spaces That's clear. Thanks. Jun18 accepted Riesz Lemma for reflexive spaces Jun18 asked Riesz Lemma for reflexive spaces Jun17 asked Extension of a linear operator Jun10 accepted Exercise on abstract integration Jun10 asked Exercise on abstract integration Jun3 comment Functional defined on the space of functions with compact support I understood your solution. I was thinking about showing that the kernel of the operator is dense in the space, but I wasn't able to do it...this should stand anyway right? Jun3 asked Functional defined on the space of functions with compact support May27 comment Exercise on isometry Yes understood what you mean. Thanks. May27 comment Exercise on isometry If I take $y_n$ to be Cauchy then $||y_n-y_m||=||Tx_n-Tx_m||=||T(x_n-x_m)||=||x_n-x_m||$ because of the isometry. Sending $m$ to infinite so we have $||x_n-x||=||Tx_n-Tx||=||y_n-y||$. Being $X$ Banach $x_n$ converge to $x \in X$ and so does $y_n \in Y$. I don't know I feel like I still miss some points... May27 asked Exercise on isometry