f not Lipschitz at [0,oo]!

evinda

Well-known member
MHB Site Helper
Hi!!! I have also an other question...
Could you explain me why $f(x)=\sqrt{x},x\geq 0$ is not Lipschitz at $[0,\infty]$??How can I show this??Do I have to use the condition $|f(x)-f(y)| \leq M|x-y|,M>0$ ,to show this??

ZaidAlyafey

Well-known member
MHB Math Helper
Re: f not Lipschitz at [0,oo]!!

Hi!!! I have also an other question...
Could you explain me why $f(x)=\sqrt{x},x\geq 0$ is not Lipschitz at $[0,\infty]$??How can I show this??Do I have to use the condition $|f(x)-f(y)| \leq M|x-y|,M>0$ ,to show this??
Hint : choose $y=0$.

ThePerfectHacker

Well-known member
Re: f not Lipschitz at [0,oo]!!

Hi!!! I have also an other question...
Could you explain me why $f(x)=\sqrt{x},x\geq 0$ is not Lipschitz at $[0,\infty]$??How can I show this??Do I have to use the condition $|f(x)-f(y)| \leq M|x-y|,M>0$ ,to show this??
If $f:[0,\infty)\to \mathbb{R}$ is Lipschitz it would mean that there is a positive $M>0$ so that for all $x,y\in [0,\infty)$ we have $|f(x)-f(y)|\leq M|x-y|$. Write it out in logic symbols:
$$\exists M>0 ~ \forall x,y\in [0,\infty), ~ |f(x)-f(y)|\leq M|x-y|$$
When you negate this statement you get,
$$\forall M>0, \exists x,y\in [0,\infty), ~ |f(x)-f(y)| > M|x-y|$$
You want to show the negated version as you are claiming $f$ is not Lipschitz.

Thus, given any $M>0$ you therefore need to find two non-negative numbers $x$ and $y$ so that $|\sqrt{x} - \sqrt{y}| > M|x-y|$.

chisigma

Well-known member
Re: f not Lipschitz at [0,oo]!!

Hi!!! I have also an other question...
Could you explain me why $f(x)=\sqrt{x},x\geq 0$ is not Lipschitz at $[0,\infty]$??How can I show this??Do I have to use the condition $|f(x)-f(y)| \leq M|x-y|,M>0$ ,to show this??
We say that the function f(x) satisfies the Lipschitz condition on the interval [a,b] if there is a constant K, independent from f and from the interval [a,b] such that for all $x_{1}$ and $x_{2}$ in [a,b] with $x_{1} \ne x_{2}$ is...

$\displaystyle |f(x_{1}) - f(x_{2})| < K\ |x_{1} - x_{2}|\ (1)$

The function $\displaystyle f(x) = \sqrt{x}$ has an umbounded derivative in x=0, so that it doesn.t satisfy the Lipschitz condition in $[0,\infty]$...

Kind regards

$\chi$ $\sigma$​

evinda

Well-known member
MHB Site Helper
Re: f not Lipschitz at [0,oo]!!

If $f:[0,\infty)\to \mathbb{R}$ is Lipschitz it would mean that there is a positive $M>0$ so that for all $x,y\in [0,\infty)$ we have $|f(x)-f(y)|\leq M|x-y|$. Write it out in logic symbols:
$$\exists M>0 ~ \forall x,y\in [0,\infty), ~ |f(x)-f(y)|\leq M|x-y|$$
When you negate this statement you get,
$$\forall M>0, \exists x,y\in [0,\infty), ~ |f(x)-f(y)| > M|x-y|$$
You want to show the negated version as you are claiming $f$ is not Lipschitz.

Thus, given any $M>0$ you therefore need to find two non-negative numbers $x$ and $y$ so that $|\sqrt{x} - \sqrt{y}| > M|x-y|$.
I understand... But..is this relation $|\sqrt{x} - \sqrt{y}| > M|x-y|$ always satisfied?

ThePerfectHacker

Well-known member
Re: f not Lipschitz at [0,oo]!!

I understand... But..is this relation $|\sqrt{x} - \sqrt{y}| > M|x-y|$ always satisfied?
No. It is not always satisfied. For example, $x=0,y=0$ will make it false. You need to show it is sometimes satisfied. You need to find some $x$ and $y$ that will do it for you.

evinda

Well-known member
MHB Site Helper
Re: f not Lipschitz at [0,oo]!!

No. It is not always satisfied. For example, $x=0,y=0$ will make it false. You need to show it is sometimes satisfied. You need to find some $x$ and $y$ that will do it for you.
Isn't this relation just satisfied for $x,y \in (0,1)$??

ThePerfectHacker

Well-known member
Re: f not Lipschitz at [0,oo]!!

Isn't this relation just satisfied for $x,y \in (0,1)$??
No. Say $M=2$ so we are saying $|\sqrt{x}-\sqrt{y}| > 2|x-y|$ for all $x,y\in(0,1)$. But that is not true, just pick $x,y=1/2$.

evinda

Well-known member
MHB Site Helper
Re: f not Lipschitz at [0,oo]!!

No. Say $M=2$ so we are saying $|\sqrt{x}-\sqrt{y}| > 2|x-y|$ for all $x,y\in(0,1)$. But that is not true, just pick $x,y=1/2$.
Aha!But..to show the negated version,don't we have to find a condition that is satisfied for each x,y?Or am I wrong?

Klaas van Aarsen

MHB Seeker
Staff member
Re: f not Lipschitz at [0,oo]!!

The typical way to disprove it, is a proof by contradiction.
First assume it is Lipschitz. That is, there is some M such that the inequality holds for every x and y.
And then find an x and y, such that the Lipschitz condition is not satisfied after all.
(Hint: pick one of the 2 as zero and the other "small enough" depending on M.)

ThePerfectHacker

Well-known member
Re: f not Lipschitz at [0,oo]!!

Aha!But..to show the negated version,don't we have to find a condition that is satisfied for each x,y?Or am I wrong?
No. You just need to find one $x$ and one $y$. That is all.

evinda

Well-known member
MHB Site Helper
Re: f not Lipschitz at [0,oo]!!

No. You just need to find one $x$ and one $y$. That is all.
So,I could pick for example $x=\frac{1}{2},y=0$ and $M=1$..Right??

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The typical way to disprove it, is a proof by contradiction.
First assume it is Lipschitz. That is, there is some M such that the inequality holds for every x and y.
And then find an x and y, such that the Lipschitz condition is not satisfied after all.
(Hint: pick one of the 2 as zero and the other "small enough" depending on M.)
I picked $y=0,x=\frac{1}{M}$ and I got $M \geq 1$.Would this be a contradiction?

Klaas van Aarsen

MHB Seeker
Staff member
Re: f not Lipschitz at [0,oo]!!

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I picked $y=0,x=\frac{1}{M}$ and I got $M \geq 1$.Would this be a contradiction?
Sorry, but no, that is not a contradiction.

Let's see what we have.
The Lipschitz condition is:
$$|f(x)-f(y)| \le M|x-y|$$

With $f(x)=\sqrt x$ and with $y=0$ this becomes:
$$|\sqrt x - \sqrt 0| \le M|x-0|$$
Since the domain is restricted to $x \ge 0$, we can simplify this to:
$$\sqrt x \le Mx$$

Can you solve it for x?

And if so, can you also find an x for which it is not true?

evinda

Well-known member
MHB Site Helper
Re: f not Lipschitz at [0,oo]!!

Let's see what we have.
The Lipschitz condition is:
$$|f(x)-f(y)| \le M|x-y|$$

With $f(x)=\sqrt x$ and with $y=0$ this becomes:
$$|\sqrt x - \sqrt 0| \le M|x-0|$$
Since the domain is restricted to $x \ge 0$, we can simplify this to:
$$\sqrt x \le Mx$$

Can you solve it for x?

And if so, can you also find an x for which it is not true?
Can we square this: $\sqrt x \le Mx$ ??

If yes,then we have: $x \le M^2x^2$ and for $x=\frac{1}{4}$,we get: $\frac{1}{4} \le \frac{M^2}{16}$..This relation does not hold for $M>2$..Could we say it like that?

ThePerfectHacker

Well-known member
Remember what you need to show. Given an $M>0$ you can find $x,y\geq 0$ such that $|\sqrt{x}-\sqrt{y}| > M|x-y|$.

I will do it for $M=1$. Choose $x=\tfrac{1}{4}$ and $y=0$, then we get,
$$\left| \sqrt{\frac{1}{4}} - \sqrt{0} \right| > 1\cdot \left| \tfrac{1}{4} - 0 \right |$$
Which is true.

Now say that $M=2$, how would you choose $x$ and $y$?

Klaas van Aarsen

MHB Seeker
Staff member
Re: f not Lipschitz at [0,oo]!!

Can we square this: $\sqrt x \le Mx$ ??
Yes, you can square this, since both sides are $\ge 0$.

If yes,then we have: $x \le M^2x^2$ and for $x=\frac{1}{4}$,we get: $\frac{1}{4} \le \frac{M^2}{16}$..This relation does not hold for $M>2$..Could we say it like that?
Wel... you still didn't solve for $x$ did you... can you?

evinda

Well-known member
MHB Site Helper
Remember what you need to show. Given an $M>0$ you can find $x,y\geq 0$ such that $|\sqrt{x}-\sqrt{y}| > M|x-y|$.

I will do it for $M=1$. Choose $x=\tfrac{1}{4}$ and $y=0$, then we get,
$$\left| \sqrt{\frac{1}{4}} - \sqrt{0} \right| > 1\cdot \left| \tfrac{1}{4} - 0 \right |$$
Which is true.

Now say that $M=2$, how would you choose $x$ and $y$?
For $M=2$ I would pick $x=\sqrt{1}{5},y=0$ ..

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Yes, you can square this, since both sides are $\ge 0$.

Wel... you still didn't solve for $x$ did you... can you?
$x^2M^2-x \ge 0 \Rightarrow x(xM^2-1) \ge 0$

Klaas van Aarsen

MHB Seeker
Staff member
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$x^2M^2-x \ge 0 \Rightarrow x(xM^2-1) \ge 0$
Let's try it like this:
\begin{array}{}
x &\le& M^2x^2 \\
\frac 1 {M^2} &\le& x \\
x &\ge& \frac 1 {M^2}
\end{array}

So let's pick $$\displaystyle x = \frac 1 {M^3}$$

evinda

Well-known member
MHB Site Helper
Let's try it like this:
\begin{array}{}
x &\le& M^2x^2 \\
\frac 1 {M^2} &\le& x \\
x &\ge& \frac 1 {M^2}
\end{array}

So let's pick $$\displaystyle x = \frac 1 {M^3}$$
Oh yes!!! This is a contradiction!!!