example of Lipschitz condition

Statement 1.

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

Proof.

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

$$\left|x^2-y^2\right|\le L\left|x-y\right|.$$

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

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|}}$$=2\left|z\right|$

$\le 2sup\{\left|z\right|:z\in (a,b)\}$

$=2\max \{\left|a\right|,\left|b\right|\}.$

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

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

as required.∎

Statement 2.

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

Proof.

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

${\displaystyle \frac{\left|f(b)-f(b-\epsilon )\right|}{\left|b-(b-\epsilon )\right|}}$$={\displaystyle \frac{b^2-{(b-\epsilon )}^2}{b-(b-\epsilon )}}$

$={\displaystyle \frac{b^2-b^2+2b\epsilon -{\epsilon }^2}{b-b+\epsilon }}$

$={\displaystyle \frac{2b\epsilon }{\epsilon }}$

$=2b.$

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

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

However, the result from the previous proof gives

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

Combining these inequalities provides

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

and the result follows by trichotomy.∎