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Let $\delta$ be an interval in $\mathbb{R}$. Recall that a function $f$ is called Lipschitz continuous on $\delta$ with Lipschitz constant $L$ if there holds $|f(x) - f(y)| \leq L|x-y|$ for all $x,y$ in $\delta$.

(a) Show that the composition of Lipschitz continuous functions is again Lipschitz continuous.

(b) Is the pointwise maximum of two Lipschitz continuous functions necessarily Lipschitz continuous?

$|f_2(f_1(y)) − f_2(f_1(x))| ≤ L2|f_1(y) − f_1(x)| ≤ L1L2|y − x|$ for part a, but i'm not sure if its right.

Anne Bauval
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Risa
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4 Answers4

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Part (a) is obvious. For (b) let $\max\{f,g\}=:h$. Then $$f(y)\leq f(x)+|f(y)-f(x)|\leq h(x)+L_f|y-x|\ ,$$ and similarly $g(y)\leq h(x)+L_g|y-x|$. It follows that $$h(y)\leq h(x)+\max\{L_f,L_g\}|y-x|\ ,$$ and by symmetry we conclude that $h$ is Lipschitz continuous with Lipschitz constant $L_h\leq \max\{L_f,L_g\}$.

Christian Blatter
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4

For part b, I think YES.

Because,

$|\max(f(x),g(x))-\max(f(y),g(y))|$

can take $4$ values:

  1. $|f(x)-f(y)|$,
  2. $|g(x)-g(y)|$,
  3. $|f(x)-g(y)|$ and
  4. $|g(x)-f(y)|$.

For the first $2$ cases, $L1$ and $L2$ are the required Lipschitz constants, where

$|f(x)-f(y)| \le L1 |x-y|$

and

$|g(x)-g(y)|\le L2 |x-y|$

Now, for the $3$rd case, wlog, assume that $x < y$.

Then we must have $f(x)\ge g(x)$, but $f(y) \le g(y)$, so by the intermediate value theorem for the continuous function $f-g$, we get a point t such that $x < t < y$, satisfying: $f(t)=g(t)$.

Now,

$$|f(x)-g(y)| \le |f(x)-f(t)|+|f(t)-g(t)|+|g(t)-g(y)|\le > L1(t-x)+L2(y-t) \le (\max(L1,L2))*(y-x)$$

The $4$th case is similarly dealt with and $\max(L1,L2)$ is your required Lipschitz constant for part b. Part a is all right, $L1*L2$ being the Lipschitz constant. Please correct me if I'm mistaken anywhere.

Angus
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Somabha Mukherjee
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  • I'm considering the case where $x$, $y$ are defined on a general metric space and we define Lipschitz continuity in terms of the general metric $d(x,y)$ rather than Euclidean distance $|x-y|$. I'm not aware of an intermediate value theorem for my case. Can I apply a modified version of this argument for my case? – snacks Jan 13 '16 at 18:32
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$g(x)$, $f(x)$ $\in$ $\mathbb{R}$ and $max\{f(x),g(x)\}=\dfrac{f(x)+g(x)+|g(x)-f(x)|}{2}\equiv h(x)$ (this can be proved in 2 cases)

We suposse that $f,g$ are Lipschitz continuos

$|f(x)-f(y)|\leq L_1|x-y|$,

$|g(x)-g(y)|\leq L_2|x-y|$,

and I will use

$|a+b|\leq |a|+|b|$

$||a|-|b||\leq |a-b|$

now

$|h(x)-h(y)|=\left|\dfrac{f(x)+g(x)+|g(x)-f(x)|}{2}-\dfrac{f(y)+g(y)+|g(y)-f(y)|}{2}\right|=\left|\dfrac{f(x)-f(y)}{2}+\dfrac{g(x)-g(y)}{2}+\dfrac{|g(x)-f(x)|-|g(y)-f(y)|}{2}\right|\leq \left|\dfrac{f(x)-f(y)}{2}\right|+\left|\dfrac{g(x)-g(y)}{2}\right|+\left|\dfrac{|g(x)-f(x)|-|g(y)-f(y)|}{2}\right|\leq \left|\dfrac{f(x)-f(y)}{2}\right|+\left|\dfrac{g(x)-g(y)}{2}\right|+\left|\dfrac{g(x)-g(y)}{2}\right|+\left|\dfrac{f(x)-f(y)}{2}\right|\leq L_1|x-y|+L_2|x-y|=(L_1+L_2)|x-y|$

Rafael Martínez
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(b) We have $\max (f,g) = (|f - g| + f + g)/2.$ It's very easy to see the sum and difference of Lipschitz functions is Lipschitz, as is any constant times a Lipschitz function. We also know that $|x|$ is Lipschitz on $\mathbb R.$ It follows that both $f-g, f+g$ are Lipschitz, hence so is $|f-g|$ by (a). Thus $|f-g| + f+g$ is Lipschitz, hence so is $(1/2)(|f-g| + f+g)$ and we're done.

zhw.
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