I have seen a proof in $\sqrt x$ is uniformly continuous

Below shows an alternative proof. Please correct me if im wrong.


For any given $\varepsilon >0$,

Let $\delta_1 = \frac{\varepsilon}{2}$, $\forall x,y \in [1,\infty)$ with $|x-y|<\delta_1$

Since $|\sqrt x + \sqrt y|\geq2$

$$|\sqrt x - \sqrt y| = \frac{|x-y|}{|\sqrt x+\sqrt y|} < |x-y| < \delta_1 = \frac{\varepsilon}{2}$$

Hence, $\sqrt x$ is uniformly continuous on $[1,\infty)$.

$\sqrt x$ is continuous on [0,1] , so $\sqrt x$ is uniformly continuous on [0,1].

So, there exist $\delta_2 > 0$ such that $\forall x,y \in [0,1], |x-y|<\delta_2$, $|\sqrt x -\sqrt y| <\frac{\varepsilon}{2}$

Let $\delta = \min{(\delta_1,\delta_2)}$

$\forall x,y \in [0,\infty)$ with $|x-y|<\delta$,

Case 1: $x,y \in [0,1]$ Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$

Case 2: $x,y \in [1,\infty)$ Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$

Case 3: $x \in [0,1] , y \in [1,\infty]$

$$|\sqrt x-\sqrt y| = |\sqrt x -1+1-\sqrt y| \leq |\sqrt x-1| + |\sqrt y -1| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}= \varepsilon$$

by applying case 1 and case 2.

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    $\begingroup$ $\frac{|x-y|}{|\sqrt x +\sqrt y|}<|x-y|$ is only true if $|\sqrt x +\sqrt y|>1$ $\endgroup$ – user223391 Apr 21 '17 at 21:43
  • $\begingroup$ Oh my, thanks for pointing it out. $\endgroup$ – Little Rookie Apr 21 '17 at 21:43
  • $\begingroup$ @ZacharySelk i have edited my proof, can u please check it? :) $\endgroup$ – Little Rookie Apr 21 '17 at 22:21
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    $\begingroup$ @Behnam i have used the theorem : If $f$ is continuous on a compact set $K \subseteq \mathbb{R}$, then $f$ is uniformly continuous on $K$. $\endgroup$ – Little Rookie Apr 21 '17 at 23:58
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    $\begingroup$ Yes, Sorry, my bad! I took it it for Lipschitzness. However, this gives an idea for the proof: On $[\delta, \infty)$ the derivative is bounded (by $(2\sqrt{\delta})^{-1}$, hence uniformly cnts there. On $[0,\delta]$ either use thrm, or directly argue that . $ \sqrt{x}-\sqrt{y} < 2\sqrt{\delta}$ Now, given a positive number, first fix delta to make the latter case be smaller than that number, then require x and y to be close enough so that on infinity side the uniform cntnty condition holds -- possible by bbddness of derivative. $\endgroup$ – Behnam Esmayli Apr 22 '17 at 6:33

Let $0 < \epsilon < 1$ and $x,y \in [0,\infty)$ with $|x-y| < \epsilon^2$.

Consider $|\sqrt{x} - \sqrt{y}| = ||\sqrt{x}| - |\sqrt{y}|| \leq |\sqrt{x} + \sqrt{y}|$.

So, $|\sqrt{x} - \sqrt{y}|^2 = |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} + \sqrt{y}| = |x-y|$, that is, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}}$ which implies that $\sqrt{\centerdot}$ is Hölder-continuous.

Finally, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}} < \sqrt{\epsilon^2} = \epsilon$.




Let $f: \;x\mapsto \sqrt{x} $ and $\epsilon>0$.

$f $ is uniformly continuous at the compact $[0,\frac 1 4] $ since it is continuous. there exist $\delta_1>0$ such that $$\forall (x,y)\in [0,\frac 1 4]^2$$

$|x-y|<\delta_1\implies|f (x)-f (y)|<\epsilon $

$f $ is uniformly continuous at $[\frac 1 4,+\infty) $ since it is lipschitzienne. $(|f (x)-f (y)|=\frac {|x-y|}{f(x)+f(y)|}\leq |x-y|)$.

there exist $\delta_2>0$ such that $$\forall x,y\geq \frac 1 4$$

Now we treat the case $x\leq \frac 14\leq y.$ $f $ is continuous at $\frac 14$, thus there exist $\delta_3>0$ such that $$|x-\frac 1 4 |<\delta_3\implies |f (x)-\frac 1 2|<\frac {\epsilon}{2} $$.

so if $x\in [0,\frac 1 4] $ and $y\in [\frac 1 4,+\infty) $ with $|x-y|<\delta_3$ then

$|x-\frac 1 4|<\delta_3$ and $|y-\frac 1 4|<\delta_3$ thus

$$|f (x)-\frac 1 2|<\frac {\epsilon}{2} $$ and $$|f (y)-\frac 1 2|<\frac {\epsilon}{2} $$

$$\implies $$

$$|f (x)-f (y)|\leq|f (x)-\frac 1 2|+|f (y)-\frac 1 2|<\epsilon $$

Now we take $$\delta=\min (\delta_1,\delta_2,\delta_3) $$

Can you finish.


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