Final answer:
To show that the convergence is uniform, we need to prove that for any epsilon greater than zero, there exists an integer N such that |fn(x) - f(x)| < epsilon for all n > N and for all x in the interval [0,1].
Step-by-step explanation:
To show that the convergence is uniform, we need to prove that for any epsilon greater than zero, there exists an integer N such that |fn(x) - f(x)| < epsilon for all n > N and for all x in the interval [0,1]. Since fn is Lipschitz continuous with a Lipschitz constant l, we can use the Lipschitz condition to obtain |fn(x) - f(x)| < l|x-y| for any x, y in [0,1]. We can choose N such that l|x-y| < epsilon, since fn converges pointwise to f, and thus the convergence is uniform.
To prove that f is Lipschitz continuous with the same Lipschitz constant l, we can use a similar argument. Let x and y be any two points in [0,1]. Then, |f(x) - f(y)| = |f(x) - fn(x) + fn(x) - fn(y) + fn(y) - f(y)| ≤ |f(x) - fn(x)| + |fn(x) - fn(y)| + |fn(y) - f(y)|. By the pointwise convergence of fn to f, we can make the first and third term arbitrarily small. And by the Lipschitz continuity of fn, we can make the second term less than l|x-y|. Thus, f is also Lipschitz continuous with a Lipschitz constant l.