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visits member for 1 year, 8 months
seen Oct 5 at 17:37

Aug
12
asked Evolutionary algorithm
Jul
2
awarded  Curious
Jun
28
accepted Exercise on isometry
Jun
20
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 $\lambda(G^{-1}(N) \cap H) \leq \lambda(G^{-1}(N))=\lambda(N)=0$.
Jun
20
asked Change of variable for Lebesegue Integral
Jun
19
accepted Theorems on continuous embedding
Jun
19
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.
Jun
19
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...
Jun
19
asked Theorems on continuous embedding
Jun
18
comment Riesz Lemma for reflexive spaces
That's clear. Thanks.
Jun
18
accepted Riesz Lemma for reflexive spaces
Jun
18
asked Riesz Lemma for reflexive spaces
Jun
17
asked Extension of a linear operator
Jun
10
accepted Exercise on abstract integration
Jun
10
asked Exercise on abstract integration
Jun
3
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?
Jun
3
asked Functional defined on the space of functions with compact support
May
27
comment Exercise on isometry
Yes understood what you mean. Thanks.
May
27
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...
May
27
asked Exercise on isometry