# example of Lipschitz condition

###### Statement 1.

Let $f(x)=x^{2}$. Then $f$ satisfies the Lipschitz condition on $[a,b]\subset\mathbb{R}$ for finite real numbers $a.

###### Proof.

We want to show that for some real constant $L$, and for all $x,y\in[a,b]$,

 $\left|{x^{2}-y^{2}}\right|\leq L\left|{x-y}\right|.$

Let $x,y\in[a,b]$. Clearly if $x=y$, the above inequality holds, so assume $x\neq y$. Since $x$ and $y$ are interchangable in the above equation, it can be assumed without loss of generality that $x.

Since $f$ is differentiable on $(a,b)$, by the mean-value theorem, there is a $z\in(x,y)$ such that

 $\frac{f(x)-f(y)}{x-y}=f^{\prime}(z),$

that is,

 $\frac{x^{2}-y^{2}}{x-y}=2z.$

Taking the modulus of both sides gives

 $\frac{\left|{x^{2}-y^{2}}\right|}{\left|{x-y}\right|}=2\left|{z}\right|.$

Finally, to find $L$ it is necessary to consider all possible values of $z$:

 $\displaystyle\frac{\left|{x^{2}-y^{2}}\right|}{\left|{x-y}\right|}$ $\displaystyle=2\left|{z}\right|$ $\displaystyle\leq 2\sup\{\left|{z}\right|\colon z\in(a,b)\}$ $\displaystyle=2\max\{\left|{a}\right|,\left|{b}\right|\}.$

Thus, for all $x,y\in[a,b]$,

 $\left|{f(x)-f(y)}\right|\leq 2\max\{\left|{a}\right|,\left|{b}\right|\}\left|{% x-y}\right|$

as required. ∎

###### Statement 2.

Additionally, $L=2\max\{\left|{a}\right|,\left|{b}\right|\}$ is the Lipschitz constant of $f$.

###### Proof.

Assume $\left|{b}\right|\geq\left|{a}\right|$, since if $\left|{b}\right|<\left|{a}\right|$, it is possible to consider $-f$ instead of $f$. This also implies that $b>0$. Let $\varepsilon>0$ be sufficiently small that $a and that higher powers of $\varepsilon$ can be ignored. Now,

 $\displaystyle\frac{\left|{f(b)-f(b-\varepsilon)}\right|}{\left|{b-(b-% \varepsilon)}\right|}$ $\displaystyle=\frac{b^{2}-(b-\varepsilon)^{2}}{b-(b-\varepsilon)}$ $\displaystyle=\frac{b^{2}-b^{2}+2b\varepsilon-\varepsilon^{2}}{b-b+\varepsilon}$ $\displaystyle=\frac{2b\varepsilon}{\varepsilon}$ $\displaystyle=2b.$

By the assumption above, $b=\max\{\left|{a}\right|,\left|{b}\right|\}$. Thus, since $b,b-\varepsilon\in[a,b]$ and by the definition of the Lipschitz condition,

 $L\geq\frac{\left|{f(b)-f(b-\varepsilon)}\right|}{\left|{b-(b-\varepsilon)}% \right|}=2\max\{\left|{a}\right|,\left|{b}\right|\}.$

However, the result from the previous proof gives

 $\frac{\left|{f(b)-f(b-\varepsilon)}\right|}{\left|{b-(b-\epsilon)}\right|}\leq L% \leq 2\max\{\left|{a}\right|,\left|{b}\right|\}.$

Combining these inequalities provides

 $2\max\{\left|{a}\right|,\left|{b}\right|\}\leq L\leq 2\max\{\left|{a}\right|,% \left|{b}\right|\},$

and the result follows by trichotomy. ∎

Title example of Lipschitz condition ExampleOfLipschitzCondition 2013-03-22 17:14:16 2013-03-22 17:14:16 me_and (17092) me_and (17092) 10 me_and (17092) Example msc 26A16