Given $I[F] := \int_0^1F^\prime(x)^2dx$, prove$|F(x_2) - F(x_1)| \leq (I[F])^\frac12 \cdot|x_2 - x_1|^\frac12$

Question

for each $ F \in C^1 [0,1]$, we define $I[F] := \int_0^1F^\prime(x)^2dx $

Using the fundamental theorem of calculus, or otherwise show that for $ F \in C^1 [0,1]$, we have $|F(x_2) - F(x_1)| \leq (I[F])^\frac12 \cdot|x_2 - x_1|^\frac12$

hint: $$|\int_a^bf(x)g(x)dx|^2 \leq(\int_a^bf(x)g(x)dx)^2\leq(\int_a^bf(x)^2dx)\cdot(\int_a^bg(x)^2dx)$$

Answer 1

$|F(x_2)-F(x_1)|=|\int_{x_1}^{x_2}F'(x) dx|=|\int_{x_1}^{x_2}F'(x) \cdot 1 dx| \le (\int_{x_1}^{x_2}F'(x)^2dx)^{1/2} \cdot (\int_{x_1}^{x_2}1^2dx)^{1/2}=(\int_{x_1}^{x_2}F'(x)^2dx)^{1/2} \cdot |x_2-x_1|^{1/2}$

and

$\int_{x_1}^{x_2}F'(x)^2dx \le I[F].$