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".

Continuity of a function is purely a local property, whereas uniform continuity is a global property that applies over the whole space.

Definition:

A real-valued function $~f~$ defined on a set $~E~$ of real numbers is said to be uniformly continuous provided for each $\epsilon~\gt ~0~$, there is a $~\delta~\gt~0~$ such that for all $~x,~x'~\in~E~$,

if$~\quad |x-x'|\lt \delta~$, then $\quad |f(x)-f(x')|\lt~\epsilon~$.

${}$

A mapping from a metric space $~(X,\rho)~$ to a metric space $~(Y,\rho)~$ is said to be uniformly continuous, provided for every $\epsilon~\gt ~0~$, there is a $~\delta~\gt~0~$ such that for all $~u,~v~\in~E~$,

if$~\quad \rho(u,~v)\lt \delta~$, then $\quad \sigma (f(x)-f(x'))\lt~\epsilon~$.

  • uniform continuous $\implies$ continuous, but converse is not true.

e.g., The function $~f(x) = \frac{1}{x}~$ is continuous on $~(0,1)~$ but not uniformly continuous.

  • The $~\delta~$ here depends on $~\epsilon~$ and on $~f~$ but that it is entirely independent of the points $~x~$ and $~y~$. In this way, uniform continuity is stronger than continuity and so it follows immediately that every uniformly continuous function is continuous.

Reference:

https://en.wikipedia.org/wiki/Uniform_continuity#Generalization_to_uniform_spaces

2113 questions
11
votes
1 answer

Uniform continuity of $f(x)=x^{2/3}\log x$ on $[0, \infty)$

Let $f$ be a real valued function on $[0,\infty)$ defined by $$f(x)= \begin{cases} x^{{2}/{3}}\log x & \text{ if } x>0 \\ 0 & \text{ if } x= 0 \end{cases}$$ Then which of the following is true? A. $f$ is discontinuous at $x=0$. B. $f$ is continuous…
user87543
6
votes
1 answer

Is the composite of an uniformly continuous sequence of functions with a bounded continuous function again uniformly continuous?

Let $\{f_n\}$ be a sequence of functions $f_n: J\to \mathbb{R}$ that converges uniformly to $f:J\to \mathbb{R}$ where $J\subseteq \mathbb{R}$ is an interval. It is clear that for a uniformly continuous function $g:\mathbb{R}\to\mathbb{R}$, the…
5
votes
2 answers

Uniform continuity on an open interval?

Suppose I want to check if $f(x)$ is uniform continuous on a bounded interval $I$ (for eg open interval $(0,1)$), given that it is continuous on $I$. How do I do that? My approach: Take $\bar{I}$, then two case can happen: Case I: If I can…
5
votes
1 answer

Show that $ x^{(1/3)}$ is uniformly continuous on $\mathbb{R}$

I've proved it when restricted to the positive and negative portions of the domain (by assuming the x and y in consideration are both positive or negative, respectively). I cannot for the life of me figure out how to prove it when x (assumed to be…
BenL
  • 981
4
votes
2 answers

uniform continuity on $[0, +\infty)$

Let $f:[0,\infty)\rightarrow R$ be a continuous function. If $\lim_{x\rightarrow+\infty} f(x)$ is finite, show that $f$ is uniformly continuous. Also, can I change "$\lim_{x\rightarrow+\infty} f(x)$ is finite" to "$f$ is bounded" and get the same…
Jason Ng
  • 365
3
votes
4 answers

How do I show the uniform continuity of $\tan^{-1}$ over $\mathbb{R}$

How do I show the uniform continuity of $\tan^{-1}$ over $\mathbb{R}$ ? I am trying to use the $\epsilon - \delta$ definition. I have just started learning this topic.
Aman Mittal
  • 2,091
2
votes
3 answers

Is the function $f(x)= {\sin x \over x}$ uniformly continuous over $\mathbb{R}$?

Is the function $$f(x)= {\sin x \over x}$$ Uniformly continuous over $R$ How do i approach this ? I need some hints.
Aman Mittal
  • 2,091
2
votes
2 answers

Uniform continuity of $\sin(x)$

I know $\sin(x)$ is uniform continuous on the non negative real line i.e on $[0,\infty)$ i tried and get it is uniformly continuous on the non positive real line i.e on $(-\infty,0]$ is then $\sin(x)$ will be uniform continuous on $\mathbb R$ please…
neha
  • 31
2
votes
1 answer

How to prove this function is not uniformly continuous?

I need to determine whether this function $f(x)=\log(2+\cos(e^x))$ is uniformly continuous on $\mathbb{R}$. I know this function is not uniformly continuous already from the graph of it, but I have no idea how to prove it formally. And I tried to…
2
votes
3 answers

If $f(x)$ is a continously differentiable function such that $f'(x)$ is unbounded, then $f(x)$ is not unifromly continuous?

Suppose $f(x)$ is a continuously differentiable function on $\mathbb{R}$ such that its derivative is unbounded. Claim: $f(x)$ is not uniformly continuous. Assume f(x) is uniformly continuous, then given $\epsilon$ we have a suitable $\delta$. Since…
2
votes
1 answer

$f(x)g(x)$ 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 it is wrong with the example $f(x)=g(x)=x$.
xldd
  • 3,485
2
votes
4 answers

Can the composition of two non-uniformly continuous functions be uniformly continuous?

I know that the composite function of two uniformly continuous functions is also uniformly continuous. But can the composition of two functions be uniformly continuous even when the functions themselves are not? What if one of them is allowed to be…
Not Euler
  • 3,079
2
votes
1 answer

Prove uniform continuity or discontinuity of function $\sqrt{x}\sin{x}$ in $(0,\infty)$

I am given a function as is mentioned in title. I want to prove it uniformly continuous or discontinuous. But I am not able to do it. Any help would be appreciated.
s.singh
  • 108
2
votes
3 answers

Show that the function $f(x) = \frac{\sin x}{x}, x >0$ is uniformly continuous.

Show that the function $f\left(x\right) = \frac{\sin x} {x}, x >0$ is uniformly continuous. Here what I have done : Let $x_1, x_2 \in \mathbb R^+$ Then, $$|f\left(x_1\right)-f\left(x_2\right) | = \left|\frac{\sin x_1} {x_1} - \frac{\sin x_2}…
Bapon Das
  • 157
2
votes
2 answers

Show that $f(x)=\frac{x}{1+|x|}$ is uniformly continuous.

I tried to use the definition and arrived this far: $|f(x)-f(y)|=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|=\frac{|x-y+x|y|-y|x||}{(1+|x|)(1+|y|)}\leq|x-y+x|y|-y|x||$. Any suggestion for ending the proof? I also tried to prove that $\frac{x}{1+x}$…
coolsv
  • 247
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