Prove that a function $g$ is Lipschitz with Lipschitz Constant without using the Mean Value Theorem.

Question

GIVEN:

• $g$ is differentiable with continuous derivative on $[a,b]$.

WANT TO SHOW

$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$

USING MEAN VALUE THEOREM

Given $x<y$ in the interval $(a,b)$, the mean value theorem states that there exists a point $\lambda \in (a,b)$ for which

$$ \frac{g(x)-g(y)}{x-y} = g'(\lambda)$$

Since $g'$ is continuous on $[a,b]$, we know that it is bounded. Thus there exists $M>0$ such that

$$g'(x) \leq M \space \space \forall x \in (a,b)$$ and in particular, $|g'(\lambda)| \leq M$. Hence,

$$\frac{|g(x)-g(y)|}{|x-y|} \leq M$$

$$\implies |g(x)-g(y)| \leq M|x-y|.$$

By the Bolzano -Weistrass Theorem there exists (since $g' \in C([a,b])$)

$$ M = \max_{x\in [a,b]} |g'(x)|.$$

Therefore,

$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$

Want to prove without using mean value theorem as it is circle work.

Answer 1

Correct me if wrong:

Assume $x \ge y$:

Then:

$g(x)-g(y) =\int^{x}_{y}g'(t)dt, $

$|g(x)-g(y)| \le \int^{x}_{y}|g'(t)|dt \le $

$M|x-y|$, where

$M = \max|g'(t)|$, $t \in [a,b].$

Similarly consider $y < x$ to complete the proof.

Answer 2

As $g \in C^1([a,b])$ you can write
$\vert g(x) - g(y) \vert = \vert \int_x^y g'(t) dt \vert \leq \int_x^y \vert g'(t) \vert dt \leq \max_{x \in [a,b]} \vert g'(x) \vert \int_x^y 1 dt = \max_{x \in [a,b]} \vert g'(x) \vert \vert x - y \vert $.