Show a function is Lipschitz

Question

Suppose a real-valued function $f : \mathbb{R} \rightarrow \mathbb{R}$ given by

$$f(x) = \begin{cases} \hfill e^{-\frac{1}{\delta^2 - x^2} + \frac{1}{\delta^2}} \hfill & \text{ $|x| < \delta$} \\ \hfill 0 \hfill & \text{ otherwise} \\ \end{cases}$$

where $\delta>0$ is any fixed real number.

Question: Is $f$ a Lipschitz function? I try to show that its derivative, which is $-\dfrac{2x}{(\delta^2 - x^2)^2}e^{-\frac{1}{\delta^2 - x^2} + \frac{1}{\delta^2}}$ is bounded and hence Lipschitz. But to no avail.

Answer 1

Begin by making a few simplifications: first, note that the $+ \frac 1 {\delta ^2}$ in the exponent plays no role in the boundedness of the derivative, being a constant. Second, note that by taking $\delta = 1$ (for an easier typing) we do not lose generality. Therefore, consider the following simplification of your function:

$$g(x) = \left\{ \begin{eqnarray} {\rm e} ^{\frac 1 {x^2 - 1}}, & -1 < x < 1 \\ 0, & \text{otherwise} \end{eqnarray} \right. .$$

You want to show that $g'$ is bounded. In order to do this, first show that it is continuous, which boils down to showing that $\lim \limits _{x \to 1, \ x < 1} g'(x) = \lim \limits _{x \to 1, \ x > 1} g'(x) = 0$, because elsewhere $g'$ is easily seen to be continuous and because $g$ is even (i.e. symmetric with respect to the $y$ axis, therefore the conclusion in $-1$ will be the one in $1$).

Getting to work, you have that

$$\lim \limits _{x \to 1, \ x < 1} g'(x) = \lim \limits _{x \to 1, \ x < 1} {\rm e} ^{\frac 1 {x^2 - 1}} \frac {-2x} {(x^2 - 1)^2} = -2 \lim \limits _{x \to 1, \ x < 1} {\rm e} ^{\frac 1 {x^2 - 1}} \frac 1 {(x^2 - 1)^2} = \cdots$$

and if you now make the change of variable $\frac 1 {x^2 - 1} = -t$, then your limit becomes

$$\cdots = -2 \lim \limits _{t \to \infty} \frac {t^2} {{\rm e} ^t} = 0$$

where on the last line you might want to use L'Hospital's rule (or you just know by heart that a polynomial over the exponential tends to $0$ whenever the variable tends to $\infty$).

Since $\lim \limits _{x \to 1, \ x > 1} g'(x) = 0$, you have that $g$ is derivable in $1$ with $g'(1) = 0$ and $g'$ continuous in $1$ (this is in fact the important conclusion). The same conclusion holds in $-1$. Since $g'$ is clearly continuous in all the other points, it is continuous everywhere.

Consider now the behaviour of $g'$ on some interval containing $[-1,1]$, for instance $[-2,2]$: the interval is compact and $g'$ is continuous on it, therefore by Weierstrass's extreme value theorem $g'$ must be bounded on $[-2,2]$. Outside of this interval, $g'$ is constant $0$, so again bounded. Therefore, $g'$ is bounded on the whole $\Bbb R$.

Answer 2

The function $f$ decreases to zero for $x\to\pm\delta$ "faster than anything", where it becomes perfectly flat.

Taking the logarithm of the derivative,

$$\ln(2x)+2\ln\left(\frac1{\delta^2-x^2}\right)+\frac1{\delta^2}-\frac1{\delta^2-x^2}=\ln(2x)+\frac1{\delta^2}+2\ln(t)-t,$$ the last term clearly dominates for increasing $t$ and the derivative goes to zero. Elsewhere, the derivative is obviously bounded.

(Actually, the derivatives of all orders are bounded.)