Is the function $\max\{f(x), 0\}$ a lipschitz continuous function?

Question

I have a function $f(x)$, which is a smooth function.

I create a new function from it $\max\{f(x), 0\}$ or $f(x)^+$. It is not smooth since the point $f(x_0)^+ = 0$ is not differentiable, but I want to know is this function Lipschitz continuous?

Answer 1

The function can be differentiable , it can also not be (see $f(x) = x$). Lipschitz-continuity depends on the domain. If the domain is a bounded and closed interval $I$, i.e. $f \in C^1(I)$, then, according to the MVT, it is Lipschitz. Then $$ \max \lbrace f(x), 0 \rbrace = \frac{1}{2}(f(x) + \lvert f(x) \rvert) $$ is also clearly Lipschitz-continuous.

So in general, if $f$ is assumed to be Lipschitz, according to the above representation, $\max \lbrace f, 0 \rbrace$ is Lipschitz, too.

If the domain is $\mathbb{R}$, then $f(x) = x^2$ is not Lipschitz and thus $\max \lbrace f(x), 0 \rbrace = f(x)$ is not Lipschitz.

Answer 2

As pointed out in the comments, if $f$ is not Lipschitz continuous to begin with, then there is little hope for $f^+$. As a concrete example, consider $f(x) := x^2$ on $\Bbb R$.

Now, suppose that $f : A \to \Bbb R$ is Lipschitz continuous, to begin with. ($A \subset \Bbb R$.) We show that $f^+$ is Lipschitz continuous as well.

Let $L > 0$ be such that $|f(x) - f(y)| \le L|x - y|$ for all $x, y \in A$.

We claim that $|f^+(x) - f^+(y)| \le L|x - y|$ for all $x, y \in A$.

Indeed, let $x, y \in A$ be arbitrary. If $f(x), f(y)$ are both positive or both non-positive, then the inequality is clearly true. Thus, assume that $f(y) \le 0 < f(x)$. In this case, we have $$|f^+(x) - f^+(y)| = f(x) \le f(x) - f(y) \le L|x - y|,$$ as desired.

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Edit: Note that I do not assume $f$ is smooth. Just that it is Lipschitz continuous.