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

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

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. – Omnomnomnom Feb 7 '14 at 13:32
• Remark: $\sqrt{}$ is certainly not defined for $x\in\mathbf R$, only for $x\geq0$. – Tom-Tom Feb 7 '14 at 13:32
• @Omnomnomnom and V.Rossetto Thanks to you both bu how can i prove it exactly – Adam Feb 7 '14 at 13:34
• Could you see it by plotting the curve of $\sqrt(x)$? Do you graphically know what "Lipschitz" mean? – Vintarel Feb 7 '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 8 '14 at 0:00

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.

• 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 7 '14 at 13:37
• What if $x<1$? Try $x = 1/4$. Is $\sqrt x < x$? – Omnomnomnom Feb 7 '14 at 13:49
• It is differentiable on $(0,1)$, but the derivative is unbounded. But yes, that is what you should look for in these problems. – Omnomnomnom Feb 7 '14 at 14:37
• See my latest edit. – Omnomnomnom Feb 7 '14 at 15:20
• – Omnomnomnom Feb 7 '14 at 15:36

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.

• @ 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 7 '14 at 14:16

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.

• @ Sepideh Bakhoda Thanks i see your point but i'm trying to see others – Adam Feb 7 '14 at 14:08

$x \geq y$ 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.

$\sqrt{}$ 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.

• Thanks but can you explain with math not with words i can't see your point ? – Adam Feb 7 '14 at 14:08