Question:
Let $f \in C^1([a;b])$, prove that $ \forall x,y \in [a;b]$ we have $|f(x)-f(y)| < || f' ||_2 \sqrt{|x-y|} $
Answer:
1- $f \in C^1([a;b]) \Rightarrow f'(\chi) $ exists and is defined: $ |f(x)-f(y)|= |\int_x ^y f'(\chi) d \chi | \leq \int_x ^y |f'(\chi)| d \chi$ by triangle inequality.
2- Now pay attention to the fact that by definition $|\int_x ^y f'(\chi) d \chi | \geq 0$ and that $\int_x ^y |f'(\chi)| d \chi \geq 0$. And as $x^2$ is an increasing function for $x \geq 0$ we have that if $|\int_x ^y f'(\chi) d \chi | \leq \int_x ^y |f'(\chi)| d \chi$ then $(\int_x ^y f'(\chi) d \chi )^2 = (|\int_x ^y f'(\chi) d \chi |)^2 \leq (\int_x ^y |f'(\chi)| d \chi)^2$
But from here I am stuck can somebody help me please?
with the help of the comment of https://math.stackexchange.com/users/4829/jose27
3- We continue by writing $(\int_x ^y |f'(\chi)| d \chi)^2 = (\int_x ^y |f'(\chi)| \cdot 1 d \chi)^2 $ thus we can identify the following inner product $\int_x ^y |f'(\chi)| \cdot 1 d \chi = \left<|f'(\chi)| | 1 \right> $
4- By CS inequality we have that $ | \left<|f'(\chi)| | 1 \right> | \leq \| f'(\chi) \|_2 \|1\|_2 = \| f'(\chi) \|_2 \sqrt{\int_x ^y 1^2 d \chi} = \| f'(\chi) \|_2 \sqrt{|x-y|}$
Q.E.D.