Let $f,g$ be Lipschitz continuous functions on $\mathbb{R}$. Let $h: \mathbb{R}^2 \to \mathbb{R}$ defined by $h(x,y) = f(x)+g(y)$. Is $h$ also Lipschitz?
I have the following:
Let $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$. SInce $f,g$ are Lipschitz, there exists $M, N$ such that for $x, y \in \mathbb{R}, |f(x) – f(y)| \leq M|x-y|$ and $|g(x) – g(y)| \leq N|x-y|$.
Now,
\begin{align}
|h(x_1,y_1) – h(x_2,y_2)| &= |f(x_1)+g(x_1) -f(x_2)-g(x_2)|\\
&= |f(x_1)-f(x_2) +g(x_1) – g(x_2)|\\
&\leq |f(x_1) – f(x_2)| + |g(x_1) – g(x_2)|\\
&\leq M|x_1 – x_2| + N|y_1-y_2|
\end{align}
How do I proceed? Or is the statement false?
Best Answer
$$ \begin{align} |f(x_1) - f(x_2)| + |g(y_1) - g(y_2)| &\leq L(f) |x_1-x_2| + L(g) |y_1-y_2| \\ &\leq \max\{L(f),L(g)\} \|(x_1,y_1)-(x_2,y_2)|_1, \end{align} $$ where $L(f)$ and $L(g)$ are the Lipschitz constants of $f$ and $g$, while $$ |(x,y)|_1=|x|+|y| $$ is an equivalent norm on $\mathbb{R}^2$.