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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?

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

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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!

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  • $\begingroup$ 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 $R_1 \cap R_2$. $\endgroup$
    – jsmath
    Commented Jul 6, 2018 at 16:24
  • $\begingroup$ @JohnSmith I have edited my question to include the proof you're seeking. I hope it clarifies your doubts. $\endgroup$
    – giobrach
    Commented Jul 6, 2018 at 17:24
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    $\begingroup$ @The answer clarified the definition. Thanks!! $\endgroup$
    – jsmath
    Commented Jul 6, 2018 at 17:43
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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.

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  • $\begingroup$ So, that seems to suggest the sum of locally Lipschitz functions is not locally Lipschitz. Or I missed something. $\endgroup$
    – jsmath
    Commented Jul 6, 2018 at 16:09
  • $\begingroup$ 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. $\endgroup$ Commented Jul 6, 2018 at 16:32
  • $\begingroup$ 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. $\endgroup$
    – jsmath
    Commented Jul 6, 2018 at 16:40
  • $\begingroup$ 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. $\endgroup$ Commented Jul 6, 2018 at 16:46

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