Question

Let {fn} be a sequence of functions defined on [0,1]. Suppose there exists a sequence of distinct numbers xn € [0, 1] such that fn(xn) = 1. Prove or disprove the following statements: a) True or false: There exists {fn} as above that converges to 0 pointwise. b) True or false: There exists {fn} as above that converges to 0 uniformly on (0,1).

Answer 1

A sequence of functions {fn} defined on [0, 1] can converge pointwise to 0, but cannot converge uniformly to 0 on [0, 1], given there is an xn where fn(xn)=1 for each n.

Step-by-step explanation:

To prove or disprove the statements, consider two types of convergence for sequences of functions: pointwise convergence and uniform convergence. A sequence of functions {fn} defined on [0, 1] that converges pointwise to 0 can indeed exist. For instance, define fn as fn(x) = 0 if x != xn and fn(xn) = 1. As n → ∞, xn can be chosen such that it does not repeat, and fn(x) for a fixed x will be 0 for sufficiently large n, except for at most one n where it might be 1. Therefore, fn converges to the 0 function pointwise.

However, the same sequence cannot converge uniformly to 0, since for every n there is an xn where fn(xn) = 1. Uniform convergence would require that the maximum difference between fn(x) and 0 on [0, 1] goes to 0 as n increases, which is not the case here. Hence, there can't be uniform convergence to 0 on [0, 1] for the given sequence of functions {fn}.