2
votes
I am missing some point about Cantor's Theorem
You are forgetting the fact that $f$ is surjective onto the power set of $A$. If $A$ is a singleton then power set of $A$ is not a sigleton. It contains the empty set also.
Existence of $\xi$ in the ...
1
vote
Accepted
Equality with metric in metric spaces
You first part of the $\impliedby$ proof has a mistake. You write
"suppose $x < y$"
but you should have written
"suppose $x \ne y$".
You've also been a little slopping elsewhere,...
1
vote
Accepted
The map $g \mapsto \phi_g$ forms an isometry from $L^q$ to $(L^p)^*$
Your second sentence is wrong and probably shows the reason for your confusion: It is the map which sends $g$ to $\varphi_g$, that is $J(g):=\varphi_g$, which constitutes the isometry $J\colon L^q\to(...
1
vote
Accepted
Isomorphism between $L^q$ and $(L^p)^*$
If $g_1,g_2 \in L^{q}$ and $\int fg_1=\int fg_2$ for all $f \in L^{p}$ then $\int_E g_1 =\int_E g_2$ for every set $E$ of finite measure. This implies that $g_1=g_2$ almost everywhere.
1
vote
Accepted
Prove this limit in the weak topology of $L^2([0,T])$
For the sake of simplicity, assume that $T = 1$.
We can prove that the result holds even if $f \in L^1$, i.e.,
\begin{equation*}
\forall f \in L^1([0,1]), \qquad \int_0^1 \sum_{i=0}^{n-1} \frac{t_{i+1}...
1
vote
1
vote
Hilbert space H is separable if and only if the dual space H* is separable
The Riesz representation theorem implements a conjugate-linear isometry
$$\Phi: H \to H^*: \xi \mapsto \langle \xi\mid -\rangle$$(assuming that inner product is linear in the second factor). In ...
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