$\sqrt{x}$ 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?

Solution 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.

Solution 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.