# sum of locally Lipschitz functions

Suppose $f$ and $g$ are locally Lipschitz functions. That is, for local regions $R1$ and $R2$, constants $M$ and $N$, we have

$|f(x) - f(y)| \leq M |x-y|$, $x, y \in R_1$,

$|g(u) - f(v)| \leq N |u-v|$, $u, v \in R_2$.

To prove the sum of $f+g$ is locally Lipschitz, one may attempt to find a constant $C$ such that

$|f(x)+g(x) - f(u)-g(u)| \leq C|x-u|, x, u \in R_1 \cap R_2$.

My question: what if $R_1 \cap R_2$ is empty?

• Do you mean $R1\times R2$ or $R1\cap R2$? Commented Jul 6, 2018 at 15:27
• I would think intersection $R1\cap R2$. Commented Jul 6, 2018 at 15:48

I assume both $f$ and $g$ are real and defined on a greater set $X$ such that $R_1,R_2 \subseteq X$. (Always remember to specify the domain and codomain of functions!)

Suppose $R_1 \cap R_2 \neq \varnothing$: then all is fine, and for all $a,b \in R_1 \cap R_2$, $$\begin{split} |(f+g)(a)-(f+g)(b)| &= |f(a) + g(a) - f(b) - g(b)| \\ &\leq |f(a) - f(b)| + |g(a) - g(b)| \\ &\leq M|a-b| + N|a-b| = (M+N)|a-b| \end{split}$$ So $C := M+N$ works.

If instead $R_1 \cap R_2 = \varnothing$, then there are no $a,b \in R_1 \cap R_2$, and thus every statement starting with "$\forall a,b \in R_1 \cap R_2$" will be true – so the above procedure may be repeated without harm. This is an example of a vacuous truth.

Addendum. I misunderstood your basic question: you are trying to prove that $f$ locally Lipschitz and $g$ locally Lipschitz implies $f+g$ locally Lipschitz. It turns out that you do not need the above subtleties to prove this.

Now, the definition of locally Lipschitz is

A function $h: X \to \mathbb R$ is said to be locally Lipschitz if for all $x \in X$ there exists an open neighborhood $\Omega_x$ and a constant $L_x >0$ such that, for all $a,b \in \Omega_x$, we have $$|h(a) - h(b)| \leq L_x|a-b|.$$

Call $\Omega_x$ and $L_x$ the open sets and the constants given by the definition for the function $f$, and $\Omega'_y$ and $L'_y$ the same for the function $g$.

Let $x$ be picked arbitrarily from $X$; then the definition applied to $f$ allows us to find $\Omega_x$ and $L_x$ such that etc. Now, for all $y \in \Omega_x$ we may similarly find $\Omega'_y$ and $L'_y$ such that etc. by the definition applied to $g$. In particular, this holds for $y=x$, so that we may find $\Omega'_x$ and $L'_x$ such that etc.

Then, for all $a,b \in \Omega_x \cap \Omega'_x$, we have that both $$|f(a) - f(b)| \leq L_x |a-b|, \qquad |g(a)-g(b)| \leq L'_x |a - b|$$ hold, and we may apply the procedure I followed above to find $$|h(a) - h(b)| \leq (L_x + L'_x) |a-b|,$$ where $h = f+g$.

Now notice that the choice of $x \in X$ is arbitrary, and so for all $x \in X$ we have found an open neighborhood $\Omega''_x$ of $X$ (namely $\Omega''_x := \Omega_x \cap \Omega'_x$, which is trivially an open neighborhood of $x$) and a constant $L''_x > 0$ (namely $L''_x = L_x + L'_x$, which is clearly a positive number) such that for all $a,b \in \Omega''_x$, $$|h(a) - h(b)| \leq L''_x |a-b|.$$

This is what we needed!

• But the case that $R_1 \cap R_2$ is empty bothers me for the proof. Because the definition of locally Lipschitz doesn't mean you have to find an inequality on $R_1 \cap R_2$. Commented Jul 6, 2018 at 16:24
• @JohnSmith I have edited my question to include the proof you're seeking. I hope it clarifies your doubts. Commented Jul 6, 2018 at 17:24
• @The answer clarified the definition. Thanks!! Commented Jul 6, 2018 at 17:43

The sum of two functions is defined on the intersection of their domains. If the intersection is empty, you cannot define the sum. For instance, the function $\log x+\log(-x)$ is undefined.

• So, that seems to suggest the sum of locally Lipschitz functions is not locally Lipschitz. Or I missed something. Commented Jul 6, 2018 at 16:09
• No. The sum of locally Lipschitz functions is locally Lipschitz on its domain, if that domain is not empty. But if the domain is empty, there is not even function. Commented Jul 6, 2018 at 16:32
• The sum of locally Lipschitz functions is locally Lipschitz on a non-empty domain. If I can't find such a non-empty domain, it seems to me that I can't prove sum of locally Lipschitz functions is locally Lipschitz. Commented Jul 6, 2018 at 16:40
• For the last time: the domain of $f+g$ is the intersection of the domains of $f$ and $g$. If that intersection is empty, $f+g$ is not defined, so it does not make sense to ask wether the sum is locally Lipschitz. Commented Jul 6, 2018 at 16:46