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30 votes
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Why do we need topological spaces?

The point Preuss makes is that we cannot find a metric $d$ on the set of function such that "$f_n \to f$ pointwise" is equivalent to "$f_n \to f$ in the metric $d$" or $d(f_n,f) \...
Henno Brandsma's user avatar
29 votes
Accepted

If $(f_n')$ converges uniformly on an interval, does $(f_n)$ converge?

If, for each $n\in\Bbb N$, $f_n(x)=n$, then you always have $f_n'(x)=0$. Therefore, $(f_n')_{n\in\Bbb N}$ converges uniformly, but there is no $x\in\Bbb R$ such that $\bigl(f_n(x)\bigr)_{n\in\Bbb N}$ ...
José Carlos Santos's user avatar
25 votes

If $(f_n')$ converges uniformly on an interval, does $(f_n)$ converge?

This might help in your thinking about this. You have your simple counterexample so the real question is what extra assumption is needed so that your attempted proof would work? It would be a ...
B. S. Thomson's user avatar
16 votes

Why do we need topological spaces?

If you consider the set of all functions $f\colon I\to \mathbb R$ on some interval $I$ (just to fix some context), then you can consider (at least) two different meanings to the convergence $f_n\to f$ ...
Ittay Weiss's user avatar
  • 80.2k
15 votes

Why do we need topological spaces?

At the end of the first chapter of Willard's text "General Topology" (1970), Willard provides the following elegantly stated motivation for the theory ... $\qquad$The process which topology ...
quasi's user avatar
  • 59.1k
12 votes
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Pointwise convergence equivalent to convergence in pointwise topology

$(\Rightarrow)$ Suppose that $f_n,f\in X^A$ are such that $f_n\to f$ in the product topology. This means that, for any finite set of points $\{a_1,a_2,\dots,a_k\}\subseteq A$ and and any choice of ...
anonymous's user avatar
  • 912
12 votes

Does pointwise convergence to a continuous function on a closed interval imply uniform convergence?

Check out $f_n(x) = nx e^{-nx}$ on $[0, 1]$.
Adayah's user avatar
  • 10.7k
12 votes
Accepted

$C^\infty$ $f_n$ converge pointwise $\implies$ f is $C^\infty$

No, it is not true. Take $f_n(x)=\sqrt{\frac1n+\left(x-\frac12\right)^2}$. Then each $f_n$ is a $C^\infty$ function. But $(f_n)_{n\in\Bbb N}$ converges uniformly to $f\colon[0,1]\longrightarrow\Bbb R$ ...
José Carlos Santos's user avatar
11 votes

Theorems similar to Dini's Theorem and Egoroff's Theorem

Here's one that I rather like that uses equicontinuity and compactness as the engine for moving from pointwise to uniform convergence. For the sake of clarity, let's recall what equicontinuity means. ...
Glitch's user avatar
  • 8,466
9 votes
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What is the difference between pointwise boundedness and boundedness?

Answering your questions in order: 1) Since $B$ is a set I think pointwise compactness here is limit point compactness, which means that every infinite set contains a limit point. In metric spaces ...
postmortes's user avatar
  • 6,323
9 votes
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Language around Pointwise convergence v. Uniform convergence

Given an $\epsilon>0$ and $x$, we can define $N_x(\epsilon)$ as the smallest $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n\geq N.$ In some sense, $N_x(\epsilon)$ is a measure of the rate ...
Thomas Andrews's user avatar
9 votes
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how can I define uniform convergence in topological spaces, and also the concept of convergent series without talking about nets?

Uniform convergence simply cannot be defined in a topological space, since it is not invariant under homeomorphisms. For instance, consider the sequence of functions $f_n:\mathbb{R}_+\to\mathbb{R}$ ...
Eric Wofsey's user avatar
8 votes
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Does $L_1$ convergence of continuous functions imply pointwise convergence?

Consider $f_n(x) = (-1)^n x^n.$ Then $f_n\in C[0,1],$ and $f_n \to 0$ in $L^1,$ but $f_n(1)$ does not converge.
zhw.'s user avatar
  • 106k
8 votes

Pointwise convergence of uniformly continuous functions to zero, but not uniformly

The another standard one is the growing steeple on $[0,1]$: $$f_n(x)=\begin{cases}n^2 x &\text{if}\;0 \leq x \leq \frac{1}{n}\\ 2n-n^2 x &\text{if}\;\frac{1}{n} \leq x \leq \frac{2}{n}\\ 0 &...
Chinnapparaj R's user avatar
8 votes
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intuition for $\lim_{n \to \infty} f_n'(x) \ne f'(x)$ (Tom Apostol's exercise $11.7.18$)

I don't have a copy of Apostol at hand, but I guess he is using this as an example of the general fact that we are not always allowed to change the order of two limit processes. Here $D$ and $n\to\...
8 votes
Accepted

Why doesn't any pointwise convergent imply uniform?

If $E$ consists of finitely many points, then what you're saying works. But generally you will be interested in infinite domains, in which case the maximum you're considering need not exist. For ...
Micah's user avatar
  • 38.2k
7 votes
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Is a continuous function necessarily the pointwise limit of a sequence of uniformly continuous functions?

If the domain of $f$ is $\Bbb R$, then the answer is yes. The proof is quite easy: let $f_n: \Bbb R \to \Bbb R$ defined by $$f_n(x)= \begin{cases} f(-n) & x<-n \\ f(x) & -n \le x \le n \\ ...
Crostul's user avatar
  • 36.9k
7 votes

how do you compute the value of $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{4n-3}$

Take the partial sums $$ \sum_{n=0}^N\frac{(-1)^{n+1}}{4n+1}=-\sum_{n=0}^N\int_0^1 (-x^4)^ndx=-\int_0^1\frac{1-(-x^4)^{N+1}}{1+x^4}dx. $$ Now, $$ \left|\int_0^1\frac{(-x^4)^{N+1}}{1+x^4}dx\right|\...
Tuvasbien's user avatar
  • 9,342
6 votes
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uniform convergence of the composition of two sequence of functions

Not even true in a simple case. Take $f_n(x)=x^2$ then $$|g_n(x)-f(x)|=\left|2\frac{x}{n}+\frac{1}{n^2}\right|$$ For $n>1$ one has: $$\Vert g_n(x)-f(x)\Vert_\infty\geq \left|2\frac{n^2}{n}+\frac{1}{...
Shashi's user avatar
  • 8,828
6 votes
Accepted

Let $f_n \to f$ pointwise on $[0,1]$. If $f_n, f$ are continuous, is it true that $\int_0^1 f_n(x) dx \to \int_0^1 f(x)dx$?

No. Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}n\sin(n\pi x)&\text{ if }x\leqslant\frac1n\\0&\text{ otherwise.}\end{...
José Carlos Santos's user avatar
6 votes

Does pointwise convergence to a continuous function on a closed interval imply uniform convergence?

Another example is $$f_n(x)=\begin{cases} n^2x&0\le x\le\frac{1}{n},\\ 2n-n^2x&\frac{1}{n}\le x\le\frac{2}{n},\\ 0&\frac{2}{n}\le x\le 1 \end{cases} $$ The graph looks like a triangle of ...
saulspatz's user avatar
  • 53.2k
6 votes
Accepted

Does $\sum^{\infty}_{n=1}xe^{-nx}$ converge uniformly on $[0,\infty)$?

Analysis First write down the truncate of the series: for $x>0, m, p \in \mathbb N^*$, we have $$ \sum_{m+1}^{m+p} x \mathrm e^{-nx} = \mathrm e^{-(m+1)x} x\sum_0^{p-1} \mathrm e^{-nx} = \mathrm e^...
xbh's user avatar
  • 8,865
6 votes
Accepted

What is the main difference between pointwise and uniform convergence as defined here?

$f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$ $f_n$ converges uniformly means that for every $c>0$ ...
Tsemo Aristide's user avatar
6 votes

What is the main difference between pointwise and uniform convergence as defined here?

Uniform convergence is actually $\mathcal L^\infty$ convergence, i.e. $$ f_n \rightrightarrows f [x \in E]\!\! \iff \!\! \sup_{x \in E} \vert f_n - f\vert(x) \to 0[n \to \infty]. $$ This is strictly ...
xbh's user avatar
  • 8,865
6 votes
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If $f_n\to f$ pointwise, $f$ is continuous and $f$ is continuous, then $f_n \to f$ uniformly.

The statement cannot be proved, since it is false. Suppose, for instance, that you define, for each $n\in\mathbb N$,$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&...
José Carlos Santos's user avatar
6 votes
Accepted

Pointwise convergence of $x^n$ in $[0,1]$

Your question has been answered in comments, and the conclusion is that the text doesn't make any sense. Quotation: "With respect to the supremum-norm the sequence converges pointwise". Yes, it ...
6 votes
Accepted

The limit of a pointwise converging sequence of polynomial is smooth

Your tags don't include complex analysis, but that approach gives a quick proof, so I'll post it. We can consider each $P_n(x)=\sum_{m=0}^{M_n}a_{mn}x^m$ as defined on all of $\mathbb C,$ simply by ...
zhw.'s user avatar
  • 106k
6 votes
Accepted

Pointwise convergence of uniformly continuous functions to zero, but not uniformly

Consider $$f_n(x)=1-\min(1,n|x-1/n|)=\begin{cases} nx&\text{ if }x<\frac{1}{n}\\2-nx&\text{ if }\frac1{n}\leq x\leq\frac{2}{n}\\0&\text{ otherwise.}\end{cases}$$ Each $f_n$ is ...
Robert Z's user avatar
  • 146k
6 votes

Pointwise convergence of uniformly continuous functions to zero, but not uniformly

Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$The sequence $(f_n)_{n\in\mathbb N}$ converges pointwise to the null ...
José Carlos Santos's user avatar
6 votes
Accepted

Does pointwise converge imply uniform convergence in some interval?

Let $(q_n)_{n\in\mathbb N}$ be an enumeration of $\mathbb Q\cap[0,1]$. For each $n\in\mathbb N$, let $f_n\colon[0,1]\longrightarrow\mathbb R$ be equal to $\chi_{\{q_n\}}$; in other words, $f_n(x)=0$, ...
José Carlos Santos's user avatar

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