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seen Dec 17 at 16:19

Dec
11
comment Show that the Open Mapping Theorem requires both spaces to be complete
But $c$ must be INDEPENDENT of $n$ which is impossible.
Dec
10
answered Show that the Open Mapping Theorem requires both spaces to be complete
Dec
10
comment Show that the Open Mapping Theorem requires both spaces to be complete
If $Y$ were complete the OMT would imply that the norms $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are equivalent on $\ell^1$ which is certainly wrong (for $x=(1,.\ldots,1,0,\ldots)$ you have $\|x\|_1=n$ but $\|x\|_\infty=1$). A more direct way to see the incompleteness: The sequence $x_n=(1,1/2,\ldots,1/n,0,\ldots)$ is $\|\cdot\|_\infty$ Cauchy in $\ell^1$ but it does not have a limit in $\ell^1$.
Dec
9
awarded  Caucus
Dec
4
answered $C(X\times Y) $separates points and vanishes nowhere
Dec
3
comment Proofchecking: Application of Banach-Alaoglu on weak converging nullsequence
All you need is that $f$ is in the weak*-closure of $\lbrace f_n: n\ge m \rbrace$.
Dec
3
comment Proofchecking: Application of Banach-Alaoglu on weak converging nullsequence
I think your proof is okay and that, indeed, the result holds for normed spaces. One has to be a bit careful because Banach-Alaoglu does NOT assert that the dual unit ball is SEQUENTIALLY weak*-compact but, as far as I can see, you do not use this.
Dec
3
comment Proofchecking: Application of Banach-Alaoglu on weak converging nullsequence
The statement in the problem is obviously wrong ($x_n=0$). Is there a typo?
Dec
3
comment Trying to understand the set $D(f, \epsilon , P)$
For continuous functions (on $P$) the set is empty.
Dec
2
comment Interpolation sequences and open mapping theorem
@David $T$ open means that $T(B)$ is open in $Y$ for every open set $B$ in $X$. For the open unit ball $B=B_X(1)$ you have $0\in T(B)$ and hence the is $r >0$ with $B_Y(r)\subseteq T(B_X(1))$. Using the homogeneity of the norms and of $T$ you get that for each $y\in Y$ there is $x\in X$ with $T(x)=y$ and $\|x\|_X \le C \|y\|_Y$ with $C=1/r$.
Dec
1
comment Function of probability measures representation
If $F$ is linear it is hardly bounded in the sense that $|F(P)|\le C$ for some $C$ (except if $F=0$). Thus Luiz' remark answers the question.
Dec
1
comment Interpolation sequences and open mapping theorem
The open mapping theorem says more: If a continuous linear operator $T:X\to Y$ is surjective then it is open. Hence there is $C>0$ such that for every $y\in Y$ with $\|y\| \le 1$ there is $x\in X$ with $T(x)=y$ and $\|x\|\le C$.
Nov
27
comment When is a metrizable topological vector space locally bounded?
The locally convex case is one of the first theorems in the theory of topological vector spaces. It is due to Kolmogorov (but nowadays a simple exercise).
Nov
27
answered Graph of weakly continuous linear operator
Nov
25
comment Is there are “sphere” associated to any topological vector space?
In normed vector spaces you have $S=\lbrace x\in X: \|x\|=1\rbrace$.
Nov
20
comment Proof of an inequality in $\mathbb{R}$
Can you do the case $\epsilon=1$? Then try $\tilde x=x \epsilon$ and $\tilde y=y/\epsilon$.
Nov
20
answered Power Series: Derivative
Nov
20
comment Theorem 2.3-2 in _Introductory Functional Analysis With Applications_ by Erwine Kryszeg
Looks like the completion of the metric of $X$, and then one has to extend the vector space structure and the norm. A shorter way is to embed $X$ in its bidual. This elegant methods needs the Hahn-Banach theorem to veify that the canonical embedding ($x\mapsto \delta_x$, where $\delta_x$ is the evaluation in $x$) is an isometry.
Nov
19
comment Tensor product of function space and vector space
For $C(X,V)$ you have to use the $\varepsilon$- or injective tensor product and for $L^1(X,V)$ the $\pi$- or projective tensor product of Grothendieck.
Nov
19
comment Spivak proof for limits, rigorous, difficult.
$y\mapsto y -L$ and $y\mapsto y+L$ are continuous.