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Questions tagged [arzela-ascoli]

The Arzela-Ascoli theorem is a fundamental result of mathematical analysis giving necessary and sufficient conditions to decide whether every sequence of a given family of real-valued continuous functions defined on a closed and bounded interval has a uniformly convergent subsequence. The main condition is the equicontinuity of the family of functions. The theorem is the basis of many proofs in mathematics. Use this tag alongside (real-analysis).

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If $(f_n)$ is uniformly limited, $f_n\rightarrow_u f$, $(g_n)$ is equicontinuous, $g_n\rightarrow_u g$, then $g_n\circ f_n\rightarrow_u g\circ f$

Let $f_n:\mathbb{R}^m\rightarrow\mathbb{R}$ and $g_n:\mathbb{R}\rightarrow\mathbb{R}$ be sequences. Suppose that: $(i)$ $(f_n)_n$ is uniformly limited and $f_n\rightarrow f$ uniformly, for a certain $...
Gary Faust's user avatar
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a sequence of characteristic functions converge uniformly near t=0, then they are equicontinuous

I encountered a problem when reading A course in probability theory by Kailai Chung: If the sequence of ch.f.'s $\{f_n\}$ converges uniformly in a neighborhood of the origin, then $\{f_n\}$ is ...
Puppet's user avatar
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Improving proof: $f_n\longrightarrow f$ pointwise, $\{f_n\}$ uniformly Lipschitz, show $f_n\overset{}{\rightrightarrows} f$ not using Arzela-Ascoli

The problem is that, for $\{f_n\}$ real-valued compactly supported sequence of functions with $f_n\longrightarrow f$ pointwise, $\{f_n\}$ uniformly Lipschitz, say $|f_n(x)-f_n(y)|\leq M|x-y|$ for any $...
CCQ's user avatar
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Direct method with integral constraint

Let $\Omega\subset\mathbb{R}^n$ be nonempty, open and bounded with $C^1$ boundary. Let $p\in[1,n)$. Let $g\in C(\mathbb{R})$ satisfy $$|g(y)|\leq C(1+|y|^q)$$ for some $C<\infty$ and some $q$ with $...
hannah2002's user avatar
2 votes
1 answer
68 views

Sequence of 1-Lipschitz functions pointwise converge

I'm asked to prove the following: Let $(X, d_x)$ and $(Y, d_y)$ be two metric spaces. Let $D \subset X$ be dense. Show the following: If $f_1, f_2, \ldots$ is a sequence of $1$-Lipschitz functions so ...
Memento Mori's user avatar
1 vote
1 answer
85 views

Prove a set is compact

Let $(X,\rho)$ be a metric compact space and $F$ is the set of isometries from $X$ to itself. For every pair $f,g\in F$, define $$d(f,g)=\sup_{x\in X} \rho(f(x),g(x)).$$ Prove that $d$ is a metric in $...
lee max's user avatar
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Is the set compact or just relatively compact?

Verify the compactness of the following set in $C[0,1]$ with metric sup $$F=\{x_{\alpha}\in C[0,1]: x_{\alpha}(t)=\sin\alpha t, > \alpha\in[1,2]\}.$$ My attempt: Consider $t_0\in [0,1]$. $\forall ...
lee max's user avatar
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is the given set (relatively) compact?

Verify the compactness of the set in $C[0,1]$ with metric sup: $$ F=\{x_{n}\in C[0,1]: x_{n}(t)=t^n, n\in\mathbb{N}\}$$ My attempt: Consider $t_0\in [0,1]$. $\forall\epsilon>0, \forall x_n\in F, \...
lee max's user avatar
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Proving $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$ is Non-Decreasing for $u \in L^q(\Omega)$

Question: Let $\Omega \subset \mathbb{R}^d$ be measurable with $|\Omega|<\infty$, and let $u: \Omega \rightarrow \mathbb{R}$ be measurable. Define $$ \Phi_u:[1, \infty) \rightarrow[0, \infty], \...
CanDoMajoringMath's user avatar
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Normality of the derived family of a normal family.

I am trying to solve a problem from complex analysis which is concerning normal families.The problem is the following: Show that,if $\mathcal F\subset \mathcal H(\Omega)$ is a normal family of ...
Kishalay Sarkar's user avatar
3 votes
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Arzela-Ascoli. Why boundedness?

I'm confused by the Arzelà–Ascoli theorem since I think a set $F\subset C(X,Y)$ ($X$ compact metric space, $Y$ Banach space) is relatively compact already if $F$ is relatively compact at any point $x$ ...
MackeyTopology's user avatar
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uniform boundedness can be relaxed to pointwise boundedness in Arzela ascoli theorem?

Let $C(X)$ be the space of continuous functions with the usual norm. $X$ be a compact metric space. The Arzela Ascolis theorem says: A subset $S$ of $C(X)$ is compact iff it is uniformly bounded and ...
MackeyTopology's user avatar
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On the conditions of Arzela Ascoli theorem

Let $X$ be a compact metric space. Let $F\subseteq C(X;\mathbb{R})$ be a family of functions. Then according to Arzela Ascoli $F$ is compact if and only if $F$ is equicontinuous at any point $t$ and $...
MackeyTopology's user avatar
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Can I recover the $L^p$ version of Arzela-Ascoli theorem from the traditional one via the following procedure?

In Brezis's Functional Analysis, theorem $4.25$, he gives a $L^p$ version of Arzela-Ascoli theorem as follows Let $\mathcal{F}$ be a bounded set in $L^p(\mathbb{R}^n)$ with $1\leq p<\infty.$ ...
Tiffany's user avatar
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Family of equicontinuous function on a compact metric space $X$ to a metric space $Y$ is uniformly equicontinuous?

If $X$, $Y$ are metric spaces, $X$ is compact and $F\subseteq Y^X$ if $F$ is equicontinuous at any point $x$ i.e. for all $x\in X$ and $\varepsilon>0$ there is a $\delta>0$ such that when $d_X(x,...
MackeyTopology's user avatar
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Compactness of integral operators

came across this exercise I haven't been able to solve. I saw a very similar exercise (to prove an integral operator with a kernel is compact), but it is just different enough, I think, to warrant its ...
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Constructing an RKHS from a Kernel

I'm reading the book "High Dimensional Statistics" by Martin Wainwright just for fun (also as preparation of my PhD in computer science/Machine Learning). In particular, I'm currently ...
John's user avatar
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On Schoen and Yau's proof of the positive mass theorem: extracting a minimal surface in the limit

I'm reading Schoen and Yau's 1979 paper on the Positive Mass theorem. I'm having trouble understanding the proof of how they extracted a minimal surface as the limit of solutions to the Plateau ...
IsomorphicBunny's user avatar
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Do we have Arzela-Ascoli theorem on the space of all continuous stochastic process?

Let $\Omega$ be a polish space and $(\Omega,\mathcal{F},\mathbb{P})$ be a complete filtered probability space with the filtration generated by the standard Brownian motion. Let the space $U:=L^2\big(\...
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Arzela-Ascoli theorem on $C([a,b];\mathcal{H})$ with $\mathcal{H}$ a Hilbert space

To give some context: I need prove that a family $\mathcal{C}:=\{y: \|y(t)-y(s)\| \leq a(t) - a(s),\ \|y(t)\| \leq b(t)\}\subseteq C([0,T];\mathcal{H})$ is compact, for appropiated $a\in\operatorname{...
The_Variational_Hunter's user avatar
1 vote
1 answer
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Class of Lipschitz Functions on the unit d-dimensional ball

Let $\mathcal{F} = \{f:\mathcal{B}_d \to \mathbb{R}\;:\; \text{f is Lipschitz}\}$, where $\mathcal{B}_d = \{x \in \mathbb{R}^d\;:\: \|x\|_2 \leq 1\}$ is the unit ball in $d$ dimension. Is the class $\...
rostader's user avatar
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Prove space of strict contractions is closed

I am given the following subset of $C([0,1])$ of continuous real-valued functions on $[0,1]$, satisfying $$ \lvert f(x) - f(y) \rvert < \lvert x - y \rvert, \quad \int_0^1 f(x)^2 dx = 1 $$ (I am ...
pongdini's user avatar
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4 votes
1 answer
261 views

How to use Arzelà-Ascoli here?

The problem is Suppose $\Omega\subset\mathbb{R}^{n}$ is a bounded open set. Consider the initial-boundary value problem $$\begin{cases} \partial_{t}u(x,t)-\Delta u(x,t)=0 & \text{in}\ \Omega\...
mio's user avatar
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Ascoli-Arzela theorem with incomplete codomain

Wikipedia (https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Compact_metric_spaces_and_compact_Hausdorff_spaces) asserts the following (specialized from Hausdorff spaces to metric ...
oggius's user avatar
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1 vote
1 answer
135 views

Compactness of Differential Operator between Sobolev and $L^p$-spaces

I was wondering under which conditions the (weak) differential operator $D: W^{k,p}(\Omega)\rightarrow L^p(\Omega)$, $u \mapsto Du$ from a Sobolev space into the underlying $L^p$-space (on some open ...
LarsB's user avatar
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0 answers
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Compactness of the immersion $C^1(\bar\Omega)\subseteq C^{\alpha}(\bar\Omega)$

Thanks to all who will answer me! (and sorry for my bad english...) I have to prove the compactness of the immersion $C^1(\bar\Omega)\subseteq C^{\alpha}(\bar\Omega)$ with (respectively) the usual ...
Mirco Cappato's user avatar
1 vote
0 answers
98 views

Family of continuous bounded functions has uniformly convergent subsequence

Let $\{f_n\} \subset C(-2,2)$ with $||f_n||_{L^\infty(-2,2)}\leq 1$. Assume for any $x,y\in (-1,1)$, there holds $|f_n(x)-f_n(y)|\leq |x-y|^\alpha $ for some $\alpha\in (0,1)$ whenever $|x-y|>\frac{...
RipCheck's user avatar
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0 votes
1 answer
270 views

Countability of R in diagonalization argument of Arzela-Ascoli theorem proof

I have an intricate issue with the diagonalization argument used in the proof of Arzela-Ascoli theorem. It goes as follows: So assume that $\scr F$ has these three properties [closed, bounded, ...
Dymista's user avatar
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1 answer
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Some problems in the application of Arzelà–Ascoli theorem

Let $\Omega$ be a bounded domain in $\mathbb{R}^d$ with a smooth boundary. Consider the sequence $\{u_n(\cdot,s)\} \subset L^2(0,T;L^2(\Omega))$ such that $\lVert u_n(\cdot,s) \rVert_{ L^2(\Omega)} $ ...
mnmn1993's user avatar
  • 413
2 votes
1 answer
59 views

Proving that $x_n(t)=\sin\left(n\frac{(t-n^2)}{n+1}\right)$ has subsequence that converges

I'm having some trouble with the following exercise: Let $x_n:[a,b]\to\mathbb R$ be a sequence of functions given by$$x_n(t)=\sin\left(n\frac{(t-n^2)}{n+1}\right)$$ Prove that $x_n$ has a subsequence ...
Eduardo Magalhães's user avatar
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1 answer
324 views

Equicontinuity and uniformly convergent subsequences

I have been trying to grasp equicontinuity lately and with it, the Arzela-Ascoli Theorem. It says that if ${f_k}$ is a sequence of functions on a compact interval that is uniformly bounded and ...
DjuroPucar's user avatar
1 vote
1 answer
96 views

Show that $\left\{x\in[0,1] \rightarrow F(x) = \int_{0}^{x} f(t)dt : f\in C([0,1])\text{ and }||f||_\infty \leq B\right\}$ is compact.

I am trying to show that $\mathscr{F}$ is compact, where $$\mathscr{F} = \left\{F(x) = \int_{0}^{x} f(t)dt\mid x \in [0,1],f\in C([0,1]) \text{ and } ||f||_\infty \leq B\right\}$$ To do so, I was ...
The Wanderer's user avatar
1 vote
0 answers
79 views

Equivalent conditions of the Arzela-Ascoli theorem

I am taking functional analysis class and we stated the Arzela-Ascoli theorem in the following way: Let $(K, d)$ be a compact metric space, $(Y, \| · \|_{Y} )$ a Banach space and $\mathcal{F} \...
tornt's user avatar
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3 votes
0 answers
86 views

Compact embedding in $C^{k,2+\alpha}$ space

Could anyone please help me understand the compact embedding of $C^{k,2+\alpha}$ space? I am reading a paper about second order elliptic pde on $C^{k,2+\alpha}$ space. Here is the link of the paper ...
Dddduuu's user avatar
  • 113
4 votes
1 answer
699 views

Arzelà-Ascoli and Compactness

Currently I'm reading baby Rudin and I've come across Arzelà-Ascoli Theorem (Theorem 7.25 in Rudin). I've read online that equicontinuity is exactly the additional property that we need to get ...
Benny's user avatar
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0 votes
1 answer
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A question related to a family of continuously differentiable functions

Consider $X=C^{1}([-1,1])$, the space of real valued $C^1$ functions defined on the interval $[-1,1]$. Define a norm on $X$ by $||f||:=\sup_{x\in[-1,1]}(|f(x)|+|f'(x)|)$. Consider the set $Y:=\{g:||g||...
neophyte's user avatar
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-1 votes
1 answer
41 views

A question related to a family of continuous functions

Let $X:=\{g:[0,1]\rightarrow \mathbb{R}\:|\:g\in C^1,|g|\leq 10, |g^{'}| \leq 10\}$. I am interested in showing that $X$ is a relatively compact subset of $C([0,1])=\{g:[0,1]\rightarrow \mathbb{R}\:|\:...
neophyte's user avatar
  • 520
2 votes
1 answer
357 views

Prove that the integral operator $T:\mathcal{C}([0,1]) \to \mathcal{C}([0,1])$ is compact

Let $T:\mathcal{C}([0,1]) \to \mathcal{C}([0,1])$ be the operator given by $(Tf)(x) = \int \limits_{0}^{x} e^y \cdot f(y) dy, \: \forall f \in \mathcal{C}([0,1])$. where $\mathcal{C}([0,1])$ is ...
Paul Joh's user avatar
  • 559
1 vote
2 answers
136 views

Some version of the theorem of Arzelà-Ascoli

I have a question concerning the theorem of Arzelà-Ascoli. Let $(f_n)_n:[0,T]\to \mathbb{R}$ be a family of functions so that $(f_n)_n$ is uniformly bounded on $[0,T]$ and $(\frac{d}{dt}f_n)_n$ is ...
user99432's user avatar
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1 vote
1 answer
104 views

$(X,d)$ separable implies uniformly equicontinuous subset of $C(X,\mathbb{R})$ is separable?

I'm trying to prove a certain class of subsets of $C(X)$ (made into a metric space, equipped with the sup-norm) are separable iff $X$ is (where $X$ is a metric space) and while one direction is easy, ...
Isky Mathews's user avatar
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0 votes
1 answer
67 views

Arzelà–Ascoli propagation theorem

Let $E$ be a metric space and $F$ be a Banach space, $A\subset E$ dense. Let $(f_{n})$ be a squence of continuous and bounden functions from $E$ to $F$ such that the restriction of $f_{n}$ to $A$ ...
Dagoberto Mares's user avatar
2 votes
1 answer
198 views

Space of Lipschitz functions is finite dimensional

Let $(X,d)$ be a compact metric space. Let $V$ be a closed subspace of $C_{\mathbb{R}}(X)$ such that every $f\in V $ is Lipschitz. Show that $V$ is finite dimensional. Hint: Show that $A_n=\{f\in V: |...
Korn's user avatar
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0 votes
0 answers
31 views

Check the definition of "pointwise bounded under $d$"

The above definition is from Munkre's topology in Section 45.I want to know whether the definition is correct?I think it contains a typo, in place of '$a$' it should be '$x$'... Please clarify this
Styles's user avatar
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-2 votes
1 answer
151 views

Uniform convergence via Arzela-Ascoli [closed]

I want to show that $u_\epsilon = -\epsilon \log\left(\frac{ e^{\frac{x}{\epsilon}} + e^{-\frac{x}{\epsilon}}}{e^{\frac{1}{\epsilon}} + e^{-\frac{1}{\epsilon}}} \right)$ converges uniformly to $1-|x|$ ...
HelloEveryone's user avatar
1 vote
0 answers
59 views

When is the compact-open topology on homomorphisms locally compact?

Let $X$ and $Y$ be topological groups. The space $\mathrm{Cont}(X,Y)$ of continuous functions will be given the compact-open topology. The subspace $\mathrm{ContHom}(X,Y)$ of continuous homomorphisms ...
Thomas Browning's user avatar
2 votes
0 answers
220 views

Set of functions with same Lipschitz constant attains a maximum

Let $E$ be the set of all functions $u : [0, 1] \to R$ such that $u(0) = 0$ and $u$ satisfies a Lipschitz condition with Lipschitz constant $1$. Define φ : E → R according to the formula: $$ \phi(u) = ...
shdwpuppy's user avatar
0 votes
1 answer
77 views

Proved that the given set is not closed in the function space $\mathcal{C}([0,1])$

The problem is actually taken from Davidson's Real analysis: Prove that the set $S= \{ F:F(x) = \int_0^x f(t)dt, ||f|| \leq 1 ,\, f\in \mathcal{C}([0,1])\}$ is not closed. This means we should find ...
Nazono Sumiko's user avatar
10 votes
2 answers
737 views

When is the compact-open topology locally compact?

Let $X$ and $Y$ be topological spaces, and consider the compact-open topology on $C(X,Y)$, which is generated by open sets of the form $$\{\text{continuous }f\colon X\to Y:f(K)\subseteq U\}\text{ for ...
Thomas Browning's user avatar
3 votes
1 answer
62 views

Extraction of pointwise convergenct subsequence using Arzela-Ascoli theorem

Let $f_n:[a, b] \rightarrow \mathbb{R}$ be a sequence of continuous functions which is uniformly bounded i.e. $||f_n||_{L^{\infty}} \leq M <\infty$ and satisfies $f_n(a)=A$ for all $n\in \mathbb{N}....
Celestina's user avatar
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1 vote
0 answers
64 views

Solution verification of a proof of the Peano existence theorem, using Arzela-Ascoli

$\newcommand{\o}{\mathcal{O}}\newcommand{\d}{\,\mathrm{d}}$I believe what I was required to show is a general version of the Peano existence theorem. Let $\o\subseteq\Bbb R\times\Bbb R^n$ be open and ...
FShrike's user avatar
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