Final answer:
A continuously differentiable function on [0,1] satisfies the Lipschitz condition because its derivative is bounded, and by applying the Mean Value Theorem, we can find a Lipschitz constant that satisfies the condition for the entire interval.
Step-by-step explanation:
Proof of the Lipschitz Condition
For a function f that is continuously differentiable on the interval [0,1], it satisfies the Lipschitz condition. The Lipschitz condition requires that there exists a constant L such that for all x, y in the interval [0,1], the inequality |f(x) - f(y)| ≤ L|x - y| holds. Since f is continuously differentiable, its derivative, f', is continuous on [0,1]. By the Extreme Value Theorem, f' must attain a maximum value M on [0,1] because it is continuous on a closed interval. Hence, for all x and y in [0,1], by the Mean Value Theorem, there exists some c between x and y such that:
f(x) - f(y) = f'(c)(x - y)
It follows that:
|f(x) - f(y)| = |f'(c)||x - y| ≤ M|x - y|
Because the derivative does not exceed M in absolute value, f satisfies the Lipschitz condition with Lipschitz constant L = M.