# Thread: sum of two lipschitz funtions

1. ## sum of two lipschitz funtions

Let f : R ->R be a function.

We say that f is Lipschitz continuous if there is some
L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.

letting f, g : R -> R be lipschitz continious funtions,
(a) i want to show that f+g is lipschitz continious too

and also if f,g are bounded lip continuos funtions
(b) then i want to show that f.g is lipschitz continous too

does anyone know how to show these?

2. Originally Posted by dopi
Let f : R ->R be a function.

We say that f is Lipschitz continuous if there is some
L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.

letting f, g : R -> R be lipschitz continious funtions,
(a) i want to show that f+g is lipschitz continious too
Proof:

Since $\displaystyle f$ and $\displaystyle g$ are Lipschitz continuous, we have that $\displaystyle |f(x) - f(y)| < \frac {L_1}2|x - y|$ for $\displaystyle L_1 >0$ and $\displaystyle |g(x) - g(y)| < \frac {L_2}2|x - y|$ for $\displaystyle L_2 > 0$

Now let $\displaystyle L = \mbox{max} \{ L_1, L_2 \}$, and note that $\displaystyle (f + g)(x) = f(x) + g(x)$, then we have:

$\displaystyle |(f + g)(x) - (f + g)(y)| = |f(x) + g(x) - f(x) - g(x)| \le |f(x) - f(y)| + |g(x) - g(y)|$ $\displaystyle < \frac {L_1}2|x - y| + \frac {L_2}2|x - y| < \frac L2|x - y| + \frac L2|x - y| = L|x - y|$

QED

you do (b), it should be similar

3. Originally Posted by dopi
Let f : R ->R be a function.

and also if f,g are bounded lip continuos funtions
(b) then i want to show that f.g is lipschitz continous too
If $\displaystyle f$ is Lipschitz then $\displaystyle |f(x)-f(y)|\leq A|x-y|$ similarly $\displaystyle g$ so $\displaystyle |g(x)-g(y)|\leq A|x-y|$. We want to show $\displaystyle |f(x)g(x)-f(y)g(y)|\leq C|x-y|$ for some $\displaystyle C$. Note that $\displaystyle |f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|\leq$$\displaystyle |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)| \leq M_1B|x-y|+M_2A|x-y| = C|x-y|$ where $\displaystyle C=M_1B+M_2A$. Q.E.D.