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Questions tagged [uniform-continuity]

For questions involving the concept of uniform continuity, that is, "the $\delta$ in the definition is independent of the considered point".

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An element of a set with a finite cover must be an element of at most two open intervals in a subcover?

Prove: If a set $A\subseteq\mathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $x\in A$, x is an element of at most two of the open ...
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2answers
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Is $ f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} $ uniformly continuous?

Is $$ f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help. Edit: I'm looking for an answer in ...
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1answer
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Problem: Is function unifomly continuous?

Is sin ( 1/x ) uniformly continuous on set (1, + infinity) ? I tried to prove that it is Lipschitz continuous but I got stuck.
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1answer
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Is Openess and closedness are preserved under uniformly continuous map? [on hold]

Question: does uniformly continuous function maps open set to open set? and does uniformly continuous function maps closed set to closed set? "I think they are false" but I couldn't able to find ...
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1answer
12 views

Is $ f \colon \mathbb{R^2} \to \mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $\mathbb{R^2}$

Is $ f \colon \mathbb{R^2} \to \mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $\mathbb{R^2} ?$ I think it is not, I tried proving it by contradiction but can't find the right $\delta$
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42 views

Prove that $f$ is well defined and is differentiable in $[0,\infty)$. [on hold]

The function is $$f(x)=\displaystyle\sum_{n\geq1} \dfrac{x}{\sqrt n \ (x+n) }$$
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1answer
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Is there a bounded rational function that is not uniformly continuous on $\mathbb{R}$?

I think the answer is no. I tried looking for one hour and it didn't work. Isn't every rational bounded function uniformly continuous ? There's nothing online about them.
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2answers
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Is $f(x) = \frac{1}{x}$ uniformly continuous on $(1, \infty)$? [on hold]

is the function $f(x) = 1/x$ uniformly continuous on $(1, \infty)?$ I know that it is not uniformly continuous on $(0, \infty)$, but now I'm restricting it even more to get rid of most of the bad ...
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2answers
35 views

Using the sequential definition of uniform continuity to show $\sin(x)$ is uniformly continuous on $\mathbb{R}$

I want to show $\sin(x)$ is uniformly continuous on $\mathbb{R}$. Let $\{a_{n}\}$ and $\{b_{n}\}$ be sequences such that $\lim_{n\to\infty}[b_{n} - a_{n}] = 0$. Then, we need to show $\lim_{n\to\infty}...
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1answer
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Is it true that $f(x) = x^{2}$ is uniformly continuous on $\mathbb{N}$?

I think that the answer is yes. If $f(x) = x^{2}$ were uniformly continuous on $\mathbb{N}$, then for every pair of sequences $\{u_{n}\}$ and $\{v_{n}\}$ satisfying $$\lim_{n\to\infty} \left( u_{n} - ...
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0answers
50 views

Consider the series $\sum_{n=1}^\infty \frac{(-1)^n}{n+x}$

Consider the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n+x}$$ Find all $x \in \mathbb{R}$ at which the series converges. Converges absolutely. Find all intervals of $\mathbb{R}$ where the series ...
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1answer
26 views

Proof that if f(a)<0 and f(b)>0 f and is continuous on [a,b] then f changes sign at some c in (a,b) Part 2

I asked a question similar to this previously but I realized that what I was trying to prove was false. Then I changed the thing I was trying to prove but it was still false. I also accepted an answer ...
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1answer
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$f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $\lim_{x\to \infty}g(x)=0$?

on $[0,\infty)$, can we prove $f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $\lim_{x\to \infty}g(x)=0$? If $\lim_{x\to \infty}g(x)=0$ is droped, then it is easy to see that ...
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1answer
41 views

Let $ f : X \to Y $ a biyective funtion and uniformly continuous. [on hold]

Let $ f : X \to Y $ a biyective funtion and Uniformly continuous.Prove that is Y is complete and $f^{-1}$ is continuous, then X is complete.
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1answer
31 views

What is the best that you can say about a function with given property

Let $E \subset \mathbb R$, and $f:E\rightarrow \mathbb R$, be a map. Set $$\Delta_{\varepsilon,a} = \{\delta>0: B(f(a),\delta)\supset f(E\cap B(a,\varepsilon))\}\ for\ \varepsilon>0,a\in E$$ $$...
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0answers
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Does unbounded and unbounded derivative imply not uniformly continuous?

Perhaps a more precise wording of the statement I'm trying to prove says something like this: Let $I$ be an interval in $\mathbb{R}$, and allow for the possibility that $I$ is unbounded. If a ...
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0answers
26 views

A uniformly continuous function on a bounded set whose image is not bounded? [duplicate]

I'm trying to find an example of a uniformly continuous function $f:X\to Y$ such that $f(X)$ is not bounded, for some bounded metric space $X$ and a metric space $Y$. The classic exercise is to show ...
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2answers
125 views

Pointwise product of uniformly continuous functions

True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Then their pointwise product $f(x)g(x)$ is uniformly continuous. I think it will be true; ...
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A property of modulus of continuity

I'm trying to show this fact, for a generic function $f:E\subset\mathbb{R}\to\mathbb{R}$: $$ \omega(\delta _1+\delta_2)\leq \omega(\delta _1)+\omega(\delta _2) $$ where $$\omega(\delta):=\sup_{x_1,...
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1answer
42 views

What is the difference between “for every $x$ there exists a $y$” and “for all $x$ there exists a $y$”?

Is there any difference between them? or are they same? I was going through the definition of uniform continuity, which uses the term "for every". Can someone explain me the difference between the two ...
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Give an example of a set $S$ whose support function is not continuous on $R^n$.

Let $A \subseteq \mathbb{R}^n$. The support function of set $A$ is defined as the following $$ S_A(x)=\sup_{y \in A} x^Ty $$ where $x \in \mathbb{R}^n$. Notice that when $A$ is bounded the function ...
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1answer
48 views

Show that a uniformly continuous function on $[0, 1)$ must be bounded.

Suppose that $f: [0, 1) \to \Bbb{R}$ is uniformly continuous. Show that there exists $N \in \Bbb{N}$ such that for every $i \in$ {$1, 2, ..., N$} and every $x, y \in [\frac{i-1}{N}, \frac{i}{N}), |f(x)...
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1answer
31 views

How to prove that this function is continuous but not uniformly continuous?

I am having some troubles solving a question on my homework sheet: Prove that $\,\,f:\Bbb{R}^n\setminus\{0\}\to \Bbb{R}^n$, defined as $$f(x) = \frac{x}{||x||}$$ is continuous but not uniformly ...
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2answers
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Why does a uniformly continuous function on [a,b] in the reals need be closed AND bounded?

I cannot come up with any counter examples as to why a function would need to be both closed and bounded to be uniformly continuous. Why is it not sufficient to just have one condition? For example, ...
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1answer
68 views

$\epsilon-\delta$ proof that $f(x) = x \sin(1/x)$ on $(0,1)$ is uniformly continuous

I have to prove $$ f: (0,1) \to [-1,1], \ \ f(x)=x\sin(1/x) $$ using $\epsilon-\delta$. Try Fix $\epsilon >0$. I have $$ \left|x \sin(1/x) - y \sin (1/y)\right| \le |\sin (1/x)| |x-y| + |y| |\...
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2answers
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Show that the support function of a bounded set is continuous.

The support function of a set $A \in \mathbb{R}^n$ is defined as the following $$ S_A(x)=\sup_{y \in A} x^Ty $$ where $x \in \mathbb{R}^n$. Show that the support function of a bounded set is ...
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1answer
22 views

Uniform continuity on a union of sets

Let $f: X \rightarrow Y$ be a function on metric spaces. Let $E_{1}, ..., E_{n}$ be a finite collection of subsets of $X$, a metric space, such that $d(x,y) >1$ whenever $x \in E_{i}$ and $y \in E_{...
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1answer
45 views

If $f:\Bbb{R}\to\Bbb{R}$ is uniformly continuous, then $f$ is bounded

I have proved : If $f$ is uniformly continuous on $(a,b)$, then $f$ is bounded on $(a,b)$. But when it extends $(a,b)$ to $\Bbb{R}$, I don't know how to continue. Would you please give some ...
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1answer
127 views

What kind of “geometric” regularity $f'^2$ gives on $f$

When solving real-analysis' problems I like to represent the functions involved and think geometrically what is going on. Today I got the following exercise : Let $f \in \mathcal{C}^1(\mathbb{R},\...
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Proving Uniform Continuity of Fourier transform (Related to measure theory)

Let $\mu$ be a finite measure on $(\mathbf{R},\mathcal{B})$. Define $\hat\mu(t)=\int{e^{-itx} d\mu(x)} , t\in \mathbf{R}$. Show that the function $\hat\mu$ is uniformly continuous on $\mathbf{R}$. $\...
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1answer
23 views

Reverse Lipschitz condition

Assume that $f:[a,b]\to\mathbb{R}_+$ is continuous, differentiable with f'(x)>0 and Lipschitz. For $y>x$, I want to understand if $$ f(y)-f(x) \ge L (y-x) $$ holds true where $L$ is a strictly ...
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Uniform Continuity Question…

There is a question on an assignment that I am currently doing and it states (a) Let $0<\theta<\frac{\pi}{2}$ and $f(x)\neq 0$ on $[0,\theta]$ where $f$ is continuous on $[0,\theta]$. Show that ...
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1answer
36 views

Bounded Derivative Test

I'm currently working on a question in an assignment, and it asks us to use the bounded derivative test, to determine whether $$f(x)=\sin\left(\frac{1}{x}\right)$$ is uniformly continuous on the ...
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Partial derivative under integral sign

My professor gave me to prove this statement: Let $A(t,x) \in C^1([0,1]\times\Bbb R, \Bbb R).$ Then $\displaystyle \frac{\partial}{\partial x}\int_0^1A(t,x)\ dt = \int_0^1\frac{\partial A}{\...
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1answer
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Why does this argument that sin(1/x) is not uniform continuous on (0,1) fail for [0,1]?

(1): I understand that $f(x) = sin(1/x)$ is continuous at every point in the open interval (0,1), but it is not uniformly continuous on this interval (e.g. take $\epsilon=2$ and set $x_n = \frac{1}{\...
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1answer
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Proof Question: Continuous Functions on Compact Domains are uniformly continuous

So, I'm going over the proof of the following theorem in Real Analysis, but I am unable to comprehend a step in the proof. The theorem is as follows: $\textbf{Theorem:}$ Let $f$ be a continuous ...
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2answers
58 views

Let $f:(a,b) \to \mathbb{R}$ to be differentiable and unbounded. Prove: $f'$ isn't bounded on $(a,b)$

Let $f:(a,b) \to \mathbb{R}$ to be unbounded and a differentiable function. Prove: $f'$ isn't bounded on $(a,b)$ My Attempt: suppose towards contradiction that f' is bounded in $(a,b)$. then ...
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2answers
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How can a continuous function $f:[0,1] ^{\mathbb{Q}}\rightarrow \mathbb{R}$ not be uniformly continuous?

If the domain $[0,1] ^{\mathbb{Q}}$ of a continuous function $f$ consists of all rationals between zero and one, inclusive, how can the function not be uniformly continuous? From my understanding, a ...
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1answer
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a problem about continuous functions on closed intervals

This is the problem: Let f be a continuous function on a finite interval [a,b]. Suppose that f (x) > $0$ for all x in [a,b]. Prove that there is an $\alpha$ > $0$ such that f (x) > $\alpha$ for ...
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0answers
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Difference between uniformly continuous, absolutely continuous and Lipschitz continuous functions? [closed]

Can somebody please explain me the difference between uniform continuity, absolutely continuity and Lipschitz continuity using examples and counterexamples?
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1answer
34 views

$f$ is uniformly continuous on $[a,b]$ and that $g$ is uniformly continuous on $g([a,b])$. Then $f\circ g$ is uniformly continuous on $[a,b]$

Suppose that $f$ is uniformly continuous on $[a,b]$ and that $g$ is uniformly continuous on $g([a,b])$ . Then $f\circ g$ is uniformly continuous on $[a,b]$. Could anyone give me a hint for answering ...
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1answer
46 views

Confusion in a Proof

The question and its proof are given below: $ \text{Page 86, Problem 12} $ $ \color{red}{\text{A function $ f $ is said to be periodic with period $ p $ if $ f(x + p) = f(x) $ for all real $ x ...
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Determine whether the function $f(x) = \sqrt {x}$ is uniformly continuous on $A = [1, +\infty)$.

Determine whether the function $f(x) = \sqrt {x}$ is uniformly continuous on $A = [1, +\infty)$. The answer of the question is given below: But I could not understand why $\sqrt{x} + \sqrt{y} &...
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0answers
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Is there any flaw in a given proof of letter (a)?

The question and its answer is given below: I prefer the above proof than the below one: Is there any flaw in the first proof?
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1answer
25 views

Is the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$. [duplicate]

The question and its answer is given below: Determine whether t the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$. but I am wondering why the $n$ is chosen like this, ...
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1answer
40 views

Determine whether the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$.

Determine whether the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$. I see that the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$. Am I correct? ...
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1answer
27 views

If f is a uniform continuous function on (a,b), then f is bounded on (a,b).

If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$. I'm trying to understand this proof given here in this link. Its the second proof given from the question in the link, ...
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0answers
31 views

Extraneous Condition in the Hypothesis?

If $\{f_n\}$ is a sequence of continuously differentiable functions on $[a,b]$, $f_n \to f$ uniformly on $[a,b[$, and there is a function $g : [a,b] \to \Bbb{R}$ such that $f'_n \to g$ uniformly on $[...
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2answers
35 views

Non uniform continuity of $f(x)=x^3$ on the interval $[10,\infty)$

How to show that $f(x)=x^3$ is not uniform continuous on the interval $[10,\infty)$? I am well aware that $f(x)^3$ is not uniform continuous on the set of real numbers but and I believe on $[10,\...
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0answers
19 views

How does uniform continuity and continuity affect the convergence of a function

Say we are working in $L^{\infty}(\mathbb{R})$ In the below cases what can we say about the convergence? 1) $f_{n},f$ are continuous and $\lim_{n\to \infty}f_{n}=f$ 2) $f_{n},f$ are uniformly ...