Linked Questions

3
votes
2answers
622 views

Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$ [duplicate]

Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$ Let $\epsilon>0$, then there exists a $\delta=\epsilon^2$ such that $|x_1-x_2|<\delta$ for all $x_1,x_2\in[0,\infty)$. $\begin{...
4
votes
1answer
671 views

Show f(x):=sqrt(x) is uniformly continuous on [0,1] [duplicate]

Show $f(x):=\sqrt{x}$ is uniformly continuous on $[0,1]$. What I did: Need to show that $\forall \varepsilon>0: \exists\delta>0$ such that $\forall x,y\in(0,1): |x-y|<\delta\Rightarrow|f(x)-...
-1
votes
3answers
346 views

Is $\sqrt{x}$ uniformly continuous on $(0,1)$? [duplicate]

Is $\sqrt{x}$ uniformly continuous on $(0,1)$? And can anybody give me a function that is uniformly continuous on some interval but $f^2$ is not?
0
votes
1answer
201 views

Prove that f(x) = √x is uniformly continuous on [0,∞). [duplicate]

Prove that $f(x) = \sqrt x$ is uniformly continuous on $[0,∞)$. take any $x,y \in [0, \infty)$. If $|x-y| < \sigma$ then $|f(x)-f(y)|=|\sqrt x - \sqrt y|= |(x-y)(x+y)/(\sqrt x + \sqrt y)| < |\...
1
vote
1answer
121 views

Uniformly continuous [duplicate]

prove the function f(x)= Seq(Abs(x)) is uniformly continuous on its domain would someone please check my work. Thank you in advance .
0
votes
1answer
144 views

Uniform Continuity of $\sqrt{x}$ [duplicate]

I want to show $\sqrt{x}$ is uniformly continuous on $[0, \infty)$. I know it is uniformly continuous on $[0,1]$ and I can show it's uniformly cts. on $[1, \infty)$, so if I choose $x=1/2$ and $y\in [...
5
votes
3answers
9k views

Techniques to prove a function is uniformly continuous

So I understand the definition of uniform continuity, but wanted some suggestions to prove that a function is or isn't uniformly continuous. I have looked ahead and have seen that if a function is ...
1
vote
2answers
5k views

Strictly increasing continuous function

Prove that any onto strictly increasing map $f: (0,1) \to (0,1)$ is continuous. Since its strictly increasing then for $x<y$ it implies that $f(x) < f(y)$. For continuity I must show that for ...
4
votes
3answers
4k views

Direct proof. Square root function uniformly continuous on $[0, \infty)$ (S.A. pp 119 4.4.8)

(http://math.stanford.edu/~ksound/Math171S10/Hw8Sol_171.pdf) Prove for all $e > 0,$ there exists $d > 0$ : for all $x, y \ge 0$, $|x - y| < d \implies |\sqrt{x} - \sqrt{y}| < e$. (a) ...
5
votes
2answers
3k views

Prove $f(x) = \sqrt x$ is uniformly continuous on $[0,\infty)$

I have seen a proof in $\sqrt x$ is uniformly continuous Below shows an alternative proof. Please correct me if im wrong. Proof: For any given $\varepsilon >0$, Let $\delta_1 = \frac{\...
2
votes
2answers
2k views

$\sqrt{x^2+1}$ uniformly continuous on (0, 1)?

$\sqrt{x^2+1}$ uniformly continuous on (0, 1)? How to deal with such problems? Please help. I know the definition of U C. but unable to handle the problem.
7
votes
2answers
668 views

Is the matrix square root uniformly continuous?

Let $\operatorname{Psym}_n$ be the cone of symmetric positive-definite matrices of size $n \times n$. How to prove the positive square root function $\sqrt{\cdot}:\operatorname{Psym}_n \to \...
2
votes
4answers
266 views

Is $1 / \sqrt{x}$ Riemann integrable on $[0,1]$?

If $\int_0^1 1 / \sqrt{x}$ Riemann integrable then using second fundamental theorem of calculus i can easily say that $\sqrt{x}$ is uniformly continuous. Basically it has one point i.e $0$ where it ...
0
votes
3answers
252 views

How to prove that $\sqrt x$ is continuous in $[0,\infty)$?

I am trying to prove that $\sqrt x$ is continuous in $[0,\infty)$. I have started writing the following proof: Given $x_0 \in [0,\infty)$ and $\epsilon > 0$. We have to show that there exists ...
2
votes
3answers
122 views

Proving $f(x)=\sqrt{x}$ is uniformly continuous.

Where did I go wrong? The text I'm using gave a different proof. Help. Attempt: Given any $\epsilon>0$. Let $\delta=\epsilon$ so that if $\vert x-y\vert<\delta$, then \begin{align} \vert x-y\...

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