$\sqrt{x}$ isn't Lipschitz function

Question

A function f such that $$ |f(x)-f(y)| \leq C|x-y| $$

for all $x$ and $y$, where $C$ is a constant independent of $x$ and $y$, is called a Lipschitz function

show that $f(x)=\sqrt{x}\hspace{3mm} \forall x \in \mathbb{R_{+}}$ isn't Lipschitz function

Indeed, there is no such constant C where $$ |\sqrt{x}-\sqrt{y}| \leq C|x-y| \hspace{4mm} \forall x,y \in \mathbb{R_{+}} $$ we have only that inequality $$ |\sqrt{x}-\sqrt{y}|\leq |\sqrt{x}|+|\sqrt{y}| $$ Am i right ?

remark for @Vintarel i plot it i don't know graphically "Lipschitz" mean? what is the big deal in the graph of the square-root function

enter image description here

in wikipedia they said

Continuous functions that are not (globally) Lipschitz continuous The function f(x) = $\sqrt{x}$ defined on [0, 1] is not Lipschitz continuous. This function becomes infinitely steep as x approaches 0 since its derivative becomes infinite. However, it is uniformly continuous as well as Hölder continuous of class $C^{0,\alpha}$, α for $α ≤ 1/2$. Reference

1] could someone explain to me this by math and not by words, please ??

2] what does "Lipschitz" mean graphically?

If your question is whether you have proven that $\sqrt x$ is not a Lipschitz function, then the answer is no, you haven't. — Ben Grossmann, Feb 07 '14 at 13:32
Remark: $\sqrt{}$ is certainly not defined for $x\in\mathbf R$, only for $x\geq0$. — Tom-Tom, Feb 07 '14 at 13:32
@Omnomnomnom and V.Rossetto Thanks to you both bu how can i prove it exactly — Adam, Feb 07 '14 at 13:34
Could you see it by plotting the curve of $\sqrt(x)$? Do you graphically know what "Lipschitz" mean? — Vintarel, Feb 07 '14 at 14:27
@Adam, brief answer : $f$ is Lipschitz if $|f(x) - f(y)|\le C|x-y|$, so you get $$\left|\frac{f(x) - f(y)}{x-y}\right|\le C,$$ you HAVE TO recognize the left hand side term as a difference quotient (or a growth of rate) of a function, or graphically the slope of the line joining $(x,f(x))$ and $(y, f(y))$.
Thus $f$ is Lipschitz if all the secant lines are of bounded slope (is it clear?) — Vintarel, Feb 08 '14 at 00:00

Ben Grossmann · Accepted Answer · 2017-07-21T21:18:51.847

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Hint: why is it not possible to find a $C$ such that $$ |\sqrt{x} - \sqrt{0}|\leq C|x-0| $$ For all $x \geq 0$?

As a general rule: Note that a differentiable function will necessarily be Lipschitz on any interval on which its derivative is bounded.

In response to the wikipedia excerpt: "This function becomes infinitely steep as $x$ approaches $0$" is another way of saying that $f'(x) \to \infty$ as $x \to 0$. If you look at slope of the tangent line at each $x$ as $x$ gets closer to $0$, those tangent lines become steeper and steeper, approaching a vertical tangent at $x = 0$.

"Graphically", we can say that a differentiable function will be Lipschitz (if and) only if it never has a vertical tangent line.

Some functions that are not Lipschitz due to an unbounded derivative: $$ f(x) = x^{1/3}\\ f(x) = x^{1/n},\quad n = 2,3,4,5,\dots $$ A more subtle example: $$ f(x) = x^2,\quad x \in \mathbb{R}\\ f(x) = \sin(x^2), \quad x \in \mathbb{R} $$ Note in these cases that although $f'(x)$ is continuous, there is no upper bound for $f'(x)$ over the domain of interest.

edited Jul 21 '17 at 21:18

answered Feb 07 '14 at 13:34

Ben Grossmann

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i can't see why please ? guz we have $|\sqrt{x}|\leq |x|$ holds true for $\forall x \in \mathbb{R}_{+}$ with $c=1$ – Adam Feb 07 '14 at 13:37
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What if $x<1$? Try $x = 1/4$. Is $\sqrt x < x$? – Ben Grossmann Feb 07 '14 at 13:49
@ Omnomnomnom so you want to see that $\sqrt{x}$ isn't differentiable over (0,1) – Adam Feb 07 '14 at 14:12
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It is differentiable on $(0,1)$, but the derivative is unbounded. But yes, that is what you should look for in these problems. – Ben Grossmann Feb 07 '14 at 14:37
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See my latest edit. – Ben Grossmann Feb 07 '14 at 15:20
please would you add "Graphically" example to your answer i like math and picture to understand things more than words Thank you in advanced – Adam Feb 07 '14 at 15:22
I can't upload pictures, but here some more examples to try. For $f(x) = x^{1/3}$, check what happens to $f'(x)$ at and near $x = 0$. – Ben Grossmann Feb 07 '14 at 15:30
@ Omnomnomnom Thanks so much do you know any website can plot graph of fucntion with tangent please – Adam Feb 07 '14 at 15:31
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Try Wolfram Alpha (click here) – Ben Grossmann Feb 07 '14 at 15:36
what about my fuction $\sqrt(x)$ how can i see thier tangents – Adam Feb 07 '14 at 15:42
@Omnomnomnom will it be okay to say that if the graph of a function $f(x)$ intersects the $x = y$ line it wont be Lipschitz(considering $y$ to be 0) ? – Abhirikshma Feb 10 '17 at 19:43
@Abhirikshma that is certainly not the case. For example, $y = \sqrt{x}$ intersects the $y = x$ line. – Ben Grossmann Feb 10 '17 at 22:18
@Omnomnomnom $y=\sqrt{x}$ is not Lipschitz, right ? – Abhirikshma Feb 11 '17 at 02:46
@Abhirikshma I'm baffled by your question. Did you read the answer you're commenting on? Did you read the question that it answers? Why are you commenting on a 2 year old question, asking me about something only tangentially relevant? – Ben Grossmann Feb 11 '17 at 13:55
for $f(x)=x^{\frac{1}{n}}, n \in \mathbb N$ is not Lipschitz, do you need to say that $n \geq 2$? – Matata Jul 21 '17 at 18:47

score 26 · Answer 2 · answered Feb 07 '14 at 13:39

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You have $${\sqrt{1/n} - \sqrt{0}\over{1/n - 0}} = {1/\sqrt{n}\over {1\over n}} = \sqrt{n}.$$ This ratio can be made as large as you like by choosing $n$ large. Therefore the square-root function fails to be Lipschitz.

answered Feb 07 '14 at 13:39

ncmathsadist

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@ ncmathsadist Thanks why make us to say the square-root function fails to be Lipschitz based on that ratio could you be more specific ?? – Adam Feb 07 '14 at 14:16

score 12 · Answer 3 · answered Feb 07 '14 at 13:50

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Suppose that $\sqrt{x}$ is a Lipschitz function, then there exists $C$ such that

$$\Big|\frac{\sqrt{y}-\sqrt{x}}{y-x}\Big| \le C$$

Now, Let $y=2x$, so

$$(\sqrt{2}-1)x^{-\frac{1}{2}}\le C$$

Letting $x→0$ gives a contradiction.

answered Feb 07 '14 at 13:50

Sepideh Bakhoda

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@ Sepideh Bakhoda Thanks i see your point but i'm trying to see others – Adam Feb 07 '14 at 14:08

score 7 · Answer 4 · edited Feb 07 '14 at 14:54

$x \geq y$ implies $\sqrt{x} \geq \sqrt{y}$ (monotonous)

Then you need $\sqrt{x} - \sqrt{y} \leq C(x-y)$ for $x \ge y$

Use $a^2 - b^2 = \left(a-b\right)\left(a+b\right)$ to divide both sides by the (positive) $\sqrt{x} - \sqrt{y}$ to get $1 \leq C(\sqrt{x} + \sqrt{y})$, or $C \geq \frac{1}{\sqrt{x} + \sqrt{y}}$. Obviously the frac diverges as $(x,y)$ approaches $(0,0)$ so there is no upper bound $C$ to satisfy the requirement.

score 2 · Answer 5 · answered Jul 13 '21 at 09:14

Showing $\sqrt {x}$ is not Lipschitz . We prove this by contradiction method. Suppose if possible $\sqrt x$ is Lipschitz function. Then there exists $C>0$ such that $|√x-√y|<C|x-y|$ for all $x,y\in \mathbb R^{+}$. $\\$ Take $x=(\frac{1}{1+C})^2$ and $y=0$. Then $|\frac{1}{1+C}|<C((\frac{1}{1+C})^2)$ $\Rightarrow C+1\leq C$ which is not possible. So $√x$ is not Lipschitz function.

score 2 · Answer 6 · answered Feb 07 '14 at 13:56

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$\sqrt{}$ is monotonous, so just assume $x \geq y$, then you can drop the absolute values and it simplifies to $1 \leq C(\sqrt{x} + \sqrt{y}$. Since you can make the sum of square roots arbitrarily small (by suitably decreasing $x$ and $y$), as soon as it's smaller than $1/C$ the inequality no longer holds.

answered Feb 07 '14 at 13:56

huh

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Thanks but can you explain with math not with words i can't see your point ? – Adam Feb 07 '14 at 14:08

$\sqrt{x}$ isn't Lipschitz function

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