$\sqrt{x}$ 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

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

$\sqrt{}$ 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.

Answer 5

$x \geq y$ 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.

Answer 6

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.