4
$\begingroup$

Is $|x|$ Lipschitz continuous? I have started studying Lipschitz continuity. I think it is Lipschitz continuous, if you take the bound M, as per definition as 2. Then it will be Lipschitz continuous at zero. Am I right?

$\endgroup$
5
$\begingroup$

You are right.

One may observe that, for any real numbers $x,y$, we have $$ ||x|-|y||\leq 1\cdot|x-y|, $$ thus $|\cdot|$ is $k$-Lipschitz continuous with $k=1$.

$\endgroup$
1
$\begingroup$

What is Lipschitz continuity? It says that $|f(x)-f(y)| \leq M|x-y|$ for some constant $M$, for all $x,y$.

If $f(x) = |x|$, then we basically have to prove that $||x|-|y|| \leq M|x-y|$ for some constant $M$. This is a known equality and can be proved as follows:

Since $x + (x-y) = y$, we have by the triangle inequality that $|x-y| \geq |y| - |x|$.

Similarly, since $y + (y-x) = x$, we have by the triangle inequality that $|y-x| \geq |x| - |y|$.

The above reduces to $|x-y| \geq ||x|-|y||$. Hence, with $M=1$ the Lipschitz condition is satisfied.

However, $f$ is not differentiable at zero, so the absolute value function provides an important example of a Lipschitz continuous non-differentiable function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.