Why is the Lipschitz Condition described using "Cones"?

Question

I have often seen the Lipschitz Condition (Lipschitz Continuity) of mathematical functions being characterized through the following analogy: A function obeys the Lipschitz Condition if each point on the function can be passed through a double cone such that the entire function stays outside the double cone:

However, when you look at the mathematical definition of the Lipschitz Condition, all I see is the above inequality which has to be satisfied for a mathematical function to the Lipschitz Condition to be obeyed.

My Question: Can someone please tell me how the analogy of the double cone is related to the mathematical inequality of the Lipschitz Condition?

Thanks!

Reference:

• https://en.wikipedia.org/wiki/Lipschitz_continuity

Answer 1

Consider the inequality $|f(x_1) - f(x_2) | \leq K |x_1 - x_2|$. If we divide both sides by $|x_1 - x_2|$, we get $$\left|\frac{f(x_1) - f(x_2)}{x_1 - x_2}\right| \leq K.$$ Let's make the additional assumption that $f$ is differentiable and $K = 1$, so that we can replace the above inequality by $$ |f'(x) | \leq 1.$$ We can alternatively write this inequality as $$ -1 \leq f'(x) \leq 1.$$ So, $f$ is a function whose slope at each point is between $-1$ and $1$. As a result, we have $f(x) \leq x + f(0)$ and $f(x) \geq -x + f(0)$. The set of points $(x,y)$ which satisfy the inequality $$ -x + f(0) \leq y \leq x + f(0) $$ appear as a double cone when shaded in the $(x,y)$-plane.

Answer 2

Given the verified inequality, I would say that the function cannot vary (i.e increase or decrease) faster than a linear (or affine) function of slope $+K$ or $-K$, hence the cone which represents these two lines of slope $+K$ and $-K$.

Answer 3

Suppose $x_1 < x_2$. The inequality $|f(x_1)-f(x_2)| \le K |x_1 - x_2| = K(x_2-x_1)$ can then be rewritten as

$$-K(x_2-x_1) + f(x_1) \le f(x_2) \le K(x_2 - x_1) + f(x_1).$$ So with $x_1$ fixed, this is a statement about how $f$ behaves to the right of $x_1$; specifically, for $x_2 > x_1$, the function value $f(x_2)$ lies between the two lines $x \mapsto K(x_2 - x_1) + f(x_1)$ and $x \mapsto -K(x_2 - x_1) + f(x_1)$. This a right-facing cone with vertex at $(x_1, f(x_1))$.

Handling the case $x_2 > x_1$ analogously will give you the left-facing cone.