Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [weak-convergence]

For questions about weak convergence, which can concern sequences in normed/ topological vectors spaces, or sequences of measures.

Please use other tags like (tag: functional-analysis) or (tag: probability-theory).

2594 questions
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If $X_n\rightarrow X$ in distribution, how to show that $\mathbb{P}(X_n=x)\rightarrow 0$ if $F$ is continuous at $x$?

We know that $$\lim_{n\rightarrow\infty}F_n(x)=F(x)$$ whenever $F$ is continuous at $x$ (where $F_n$ and $F$ are the corresponding distribution functions), but it should also be true that $$\lim_{n\rightarrow\infty}\mathbb{P}(X_n=x)=0.$$ I tried…
Mau314
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4
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2 answers

Weak*-convergence to 0 on L^\infty and convergence almost everywhere

I am stuck with something standard... Let $f_n \in L^\infty(\mathbb{R}^d)\cap L^1(\mathbb{R}^d)$, $n\geq1$, be such that $$ \sup_{n\geq1} \|f_n\|_{L^\infty(\mathbb{R}^d)}<\infty $$ and $$ \lim_{n\to\infty} \int_{\mathbb{R}^d}|f_n(x)-f(x)|…
Dmitri
  • 231
3
votes
1 answer

Weak convergence of $\sqrt{n}e_n$ in Hilbert space

If $\{e_n| n \in \mathbb{N} \}$ is an orthonormal basis for a Hilbert space $\mathcal{H}$, then it would seem that $\langle x,\sqrt{n} e_n \rangle \to 0$ for all $x \in \mathcal{H}$, but the Banach-Steinhaus theorem implies that weakly convergent…
Lukas Rollier
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2
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1 answer

$f \in L^1$ differentiable almost everywhere with $f' \in L^1$ implies $f$ has weak derivative $f'$?

I got a function $f:[a,b] \to \mathbb{R}$ where $f \in L^1(a,b)$. It is differentiable a.e. with derivative $f' \in L^1(a,b)$. How can I show that $f$ has a weak derivative? I know that $\int f\varphi'$ and $-\int f' \varphi$ both make sense but…
C_Guy
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2
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0 answers

Taking the limit in disctretized weak forms of PDEs

Let us consider a basic one-dimensional Boundary Value Problem of the following form: given function $f : [0,1] \to\mathbb R$, the unknown solution $u : [0,1] \to\mathbb R$ satisfies $$ -u''(x)=f(x),\quad x \in(0,1),\quad u(0) = 0,\quad…
pluton
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2
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1 answer

Question on weak-star convergence in $L^\infty(0,T;L^2(\Omega))$

Could you please tell me what does it mean $u_n \rightarrow u$ weak-star in $L^\infty(0,T;L^2(\Omega))$ ?
Igor
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2
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0 answers

weak convergence in space of bounded functions

Let $B[0,1]$ be the space of bounded functions with supremum norm. Let $\{f_n\}\subset B[0,1]$, $f_n(x)=nx$ for $x\in[0,1/n]$ and $f_n(x)=0$ otherwise. Is this sequence weakly convergent to zero function in $B[0,1]$ ? My try is following: We want…
elliptic
  • 376
2
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0 answers

weakly converges distribution sense implies uniform convergence

Given a set of equicontinuous and uniformly bounded functions $\{f_n, f\}$ defined over an open connected $\Omega \in \mathbb{R}^n$. Suppose $(f_n)_n$ weakly converges to some $f$ in distribution sense. That is, assume $$f_n \rightharpoonup f…
user31899
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1
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Cauchy property of weak convergence

I found in the measure and integration theory book from Bauer (Remark 6, §20) that if $\mu$ is a finite measure, then the weak convergence of a sequence $(f_n)$ is equivalent to $$ \lim_{n,m \to \infty} \mu(\{ |f_m-f_n| \geq \alpha \})=0 $$ for all…
Adam
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vote
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weak convergence example

An example in Billingsley's book on weak convergence of measures is like this: Let $P_n$ be point mass at $x_n$ and $P$ be point mass at $x$. $P_n \Rightarrow P$ is seen to be equivalent to $x_n \to x$. Let $A = \{x_2,x_4,x_6,\ldots\}$. Weak…
user24367
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1
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0 answers

Almost sure convergence of cdf implies uniform convergence for multivariate random vectors

Let $X_1, X_2, \dots \in \mathbb{R}^d$ be random vectors, each with cdf $F_n$. Let $F$ denote the cdf of another random vector $X$. Suppose they are all continuous w.r.t. Lebesgue measure for now. I know that when $d=1$, $F_n(x) \to F(x)$ implies…
statstats
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1
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Does almost sure convergence of parameters of a parametric probability distribution imply pointwise convergence of the characteristic function?

Consider a family of distributions $\mathcal{P}=\{P_{\theta}: \theta \in \mathbb{R}^d\}$, with identifiable parameterization. If a sequence of estimators $\hat{\theta}_n \overset{a.s.}{\to} \theta_0$, does this necessarily imply a.s. convergence of…
statstats
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Does $\max{\{ f_{1}(X_{1,t}),f_{2}(X_{2,t})\}} \overset{d}{\to} \max {\{ f_{1}(X_{1,\infty}),f_{2}(X_{2,\infty})\}}$ when $f_{i}$ is continuous?

Let $X_{i,t}$ be a random variable. Support $X_{i,t}$ converges to $X_{i,\infty}$ in distribution, whose the distribution is $f_{i,\infty}$. Let $Y_{t} = \max {\{f_{i,\infty} (X_{i,t}) : \forall i=1,\ldots,n\}}$ and $Y_{\infty} = \max…
王冠倫
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Does weak convergence imply almost sure continuity of mapping

Consider standard Borel spaces $\mathcal{X}, \mathcal{Y}$ and $[0,1]$ with random variables $X, Y, U$ and joint distribution $\mathbb{P}(X, Y, U)$, where the marginal $\mathbb{P}(U) = \lambda$, i.e., $U$ has a uniform distribution on $[0,1]$. Also,…
pabk
  • 133
1
vote
1 answer

Variant of relative compactness implying variant of tightness

Let $\mu_{n, \theta}$ be a family of probability measures on a Polish space for $n \in \mathbb{N}$ and $\theta \in \Theta$. Assume that for every $\theta_n \subseteq \Theta$, the sequence $\mu_{n, \theta_n}$ has a weakly convergent sub-sequence. I'm…
Lundborg
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