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- #1

- Apr 13, 2013

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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??

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- Thread starter
- #1

- Apr 13, 2013

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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??

- Jan 17, 2013

- 1,667

Hint : choose $y=0$.

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??

- Jan 26, 2012

- 236

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:

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??

$$ \exists M>0 ~ \forall x,y\in [0,\infty), ~ |f(x)-f(y)|\leq M|x-y| $$

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 $|\sqrt{x} - \sqrt{y}| > M|x-y|$.

- Feb 13, 2012

- 1,704

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

$\chi$ $\sigma$

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

$\chi$ $\sigma$

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- #5

- Apr 13, 2013

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I understand... But..is this relation $|\sqrt{x} - \sqrt{y}| > M|x-y|$ always satisfied?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 $|\sqrt{x} - \sqrt{y}| > M|x-y|$.

- Jan 26, 2012

- 236

No. It is not always satisfied. For example, $x=0,y=0$ will make it false. You need to show it isI understand... But..is this relation $|\sqrt{x} - \sqrt{y}| > M|x-y|$ always satisfied?

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- #7

- Apr 13, 2013

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

- Jan 26, 2012

- 236

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 $x,y=1/2$.Isn't this relation just satisfied for $x,y \in (0,1)$??

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- #9

- Apr 13, 2013

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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. 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 $x,y=1/2$.

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- #10

- Mar 5, 2012

- 9,591

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.)

- Jan 26, 2012

- 236

No. You just need to find one $x$ and one $y$. That is all.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?

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- #12

- Apr 13, 2013

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

- - - Updated - - -

I picked $y=0,x=\frac{1}{M}$ and I got $M \geq 1$.Would this be a contradiction?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.)

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- #13

- Mar 5, 2012

- 9,591

Sorry, but no, that is not a contradiction.- - - Updated - - -

I picked $y=0,x=\frac{1}{M}$ and I got $M \geq 1$.Would this be 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

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- #14

- Apr 13, 2013

- 3,844

Can we square this: $\sqrt x \le Mx$ ??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 isnottrue?

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?

- Jan 26, 2012

- 236

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 $y$?

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- #16

- Mar 5, 2012

- 9,591

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

Wel... you still didn't solve for $x$ did you... can you?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?

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- #17

- Apr 13, 2013

- 3,844

For $M=2$ I would pick $x=\sqrt{1}{5},y=0$ ..

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 $y$?

- - - Updated - - -

$x^2M^2-x \ge 0 \Rightarrow x(xM^2-1) \ge 0$Yes, you can square this, since both sides are $\ge 0$.

Wel... you still didn't solve for $x$ did you... can you?

- Admin
- #18

- Mar 5, 2012

- 9,591

Let's try it like this:- - - Updated - - -

$x^2M^2-x \ge 0 \Rightarrow x(xM^2-1) \ge 0$

\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 \(\displaystyle x = \frac 1 {M^3}\)

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- #19

- Apr 13, 2013

- 3,844

Oh yes!!! This is a contradiction!!!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 \(\displaystyle x = \frac 1 {M^3}\)