Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [weak-topology]

Let $X=(X,\tau)$ be a topological vector space whose continuous dual $X^$ separates points (i.e., is T2). The weak topology $\tau_w$ on $X$ is defined to be the coarsest/weakest topology (that is, the topology with the fewest open sets) under which each element of $X^$ remains continuous on $X$.

Weak topology is an alternative term for certain initial topologies, often on topological vector spaces or spaces of linear operators, for instance on a Hilbert space.

538 questions
46
votes
0 answers

Regarding metrizability of weak/weak* topology and separability of Banach spaces.

Let $X$ be a Banach space, $X^*$ its dual, $\mathcal{B}$ the unit ball of $X$ and $\mathcal{B}^*$ the unit ball of $X^*$. The following result is well-known Theorem. If $X$ is separable, then $\mathcal{B}^*$ endowed with the weak*-topology is…
user160185
27
votes
1 answer

Prove: The weak closure of the unit sphere is the unit ball.

I want to prove that in an infinite dimensional normed space $X$, the weak closure of the unit sphere $S=\{ x\in X : \| x \| = 1 \}$ is the unit ball $B=\{ x\in X : \| x \| \leq 1 \}$. $\\$ Here is my attempt with what I know: I know that the weak…
Peter
  • 379
20
votes
2 answers

Weak topology on an infinite-dimensional normed vector space is not metrizable

I've been pondering over this problem for a while now, but I can't come up with a proof or even a useful approach... Let $X$ be am infinite-dimensional normed vector space over $\mathbb{K}$ (that is either $\mathbb{R}$ or $\mathbb{C}$). Then the…
Amarus
  • 1,797
14
votes
2 answers

Why are weak topologies useful in functional analysis?

I've been reading through chapter 3, Rudin's Functional Analysis, and an important point is the one of weak topology. From the theorems it seems to me weak topologies are somehow the result of introducing a topology of a vector space by using the…
user8469759
  • 5,591
  • 4
  • 20
  • 54
9
votes
1 answer

Are weakly compact sets bounded?

Let $X$ be a Hausdorff locally convex topological vector space, and let $X'$ denote its topological dual, that is, the vector space of all continuous linear functionals on $X$. If $A$ is a weakly compact subset of $X$, that is, if $A$ is…
serenus
  • 945
9
votes
2 answers

If $Y\subset X$ are Banach spaces such that $Y$ is dense in $X$, is it true that $X'$ is dense in $Y'$?

If $Y$ is a dense subspace of a Banach space $(X,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ is a Banach space such that the inclusion from $(Y,\|\cdot\|_2)$ into $(X,\|\cdot\|_1)$ is continuous, then it is well defined, linear, injective, and continuous in…
Bob
  • 5,995
9
votes
2 answers

Brezis Exercise 3.9

3.9 Let $E$ be a Banach space; let $M\subset E$ be a linear subspace, and let $f_{0} \in E^{\star}$. Prove that there exists some $g_{0} \in M^{\perp}$ such that $$ \inf_{g\in M^{\perp}}\lVert f_{0}-g\rVert=\lVert f_{0}-g_{0}\rVert. $$ Two methods…
D18938394
  • 1,260
8
votes
2 answers

Prove that every reflexive Banach space is weakly (sequentially) complete.

Here is the question: A sequence $(x_{n})$ in a normed linear space $X$ is weakly Cauchy if $(Tx_{n})$ is a Cauchy sequence for every $T \in X^*.$ The space $X$ is weakly (sequentially) complete if every weakly Cauchy sequence in $X$ is weakly…
user591668
8
votes
2 answers

A weakly convergent sequence in a compact set, is strongly convegnet

Let $E$ be a Banach space, and $K \subset E$, compact set for the strong topology. And let $(x_n)_n$ converges for the weak topology $\sigma(E,E^*)$ to $x$. Why $(x_n)_n$ converges for the strong topology ? My idea : Since $K$ is a compact set for…
BrianTag
  • 1,493
  • 8
  • 20
8
votes
1 answer

Weak operator topology - the operator multiplication is not a continuous function

A Hilbert space $ \mathscr l^2 $ is defined as a space with the scalar product $ (x,y)=\sum_{i=1}^\infty x_iy_i$ over $ \mathbb R $, where $x_i$ and $y_i$ are sequences. Then there is a space $L(\mathscr l^2)$, which is a space of all bounded…
Milena
  • 315
7
votes
1 answer

If $f_n \rightarrow f$ weakly in $L^p$, then $\sqrt{f_n} \rightarrow \sqrt{f}$ weakly in $L^{2p}$?

Suppose $||f_n||_{L^p(\Omega)} \leq C$, where $\Omega$ is a bounded set in $\mathbb{R}^n$. Moreover, $f_n \geq 0$. Using weak compactness, we know that there exists a subsequence $\{f_{n_k} \}$ such that $f_{n_k} \rightarrow f$ weakly in…
inhonorof Lagrange
  • 95
7
votes
2 answers

Is it necessary to consider the case $p=1$ separately?

Let $p \in [1, \infty]$. Let $f \in L^p_{\text{loc}} (\mathbb R)$ be $T$-periodic, i.e., $f(x+T) = f(x)$ a.e. $x \in \mathbb R$. Let $$ \bar f := \frac{1}{T} \int_0^T f (t) \, dt. $$ We define a sequence $(u_n) \subset L^p(0, 1)$ by $u_n (x) :=…
Analyst
  • 6,351
7
votes
1 answer

On the definition of weak and weak-* topologies

I have been studying topological vector spaces, and despite going over numerous resources, the definitions of weak and weak-* topologies have been causing me some confusion. I am having trouble visualizing and understanding these topologies. Suppose…
CBBAM
  • 7,149
7
votes
2 answers

Weak convergence in the Hilbert cube

I'm very curious about the following problem: How can I show that in the Hilbert cube defined as $$C=\{x=(x_1,x_2,\dots) \in l^p: |x_n|\leq \frac{1}{n}\,\,\, \forall n \in \mathbb{N}\}, 1\leq p < \infty$$ the weak convergence implies strong…
Victor
  • 163
  • 5
  • 17
7
votes
2 answers

In "Analyse fonctionnelle" of Brezis, in chapter III why do we need Banach spaces ? (especially for Kakutani's theorem)

In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I…
Peter
  • 1,025
1
2 3
…
35 36