Proving Uniform Convergence of {fn} to f When f is Continuous

[e(ho0n3]

Homework Statement
Say we have a sequence of functions {fn: [a,b] -> R} such that each fn is nondecreasing. Suppose that {fn} converges point-wise to f. Prove that if f is continuous, then {fn} converges uniformly to f.

The attempt at a solution
Fix an x in [a,b] and let e > 0. Then we can find a d such that if |y - x| < d, then |f(y) - f(x)| < e. Now fix a y such that |y - x| < d. Then there is an N such that |f(y) - fn(y)| < e for all n > N. And so we have

|f(x) - fn(x)| <= |f(x) - f(y)| + |f(y) - fn(y)| + |fn(y) - fn(x)|.

The first two terms on the RHS can be made arbitrarily small, but not the last one. I haven't used the fact that fn or that f is nondecreasing, but I don't understand how this would come into play. Any tips?

 

[mighty2000]

Thank you for your question. Your attempt at a solution is on the right track, but there are a few things that need to be clarified. First, let's define what it means for a sequence of functions to converge point-wise. This means that for every x in [a,b], the sequence {fn(x)} converges to f(x) as n goes to infinity. In other words, for every x, we can find an N such that for all n > N, |f(x) - fn(x)| < e. This is slightly different from what you have written, where you fix an x and then choose a y close to x.

Now, let's consider the sequence of functions {fn} and their point-wise limit f. We want to show that this sequence converges uniformly to f, which means that for every e > 0, we can find an N such that for all n > N, |f(x) - fn(x)| < e for all x in [a,b]. This is slightly different from what you have written, where you fix an x and then choose a y close to x.

To prove this, we will use the fact that f is continuous. This means that for every x in [a,b], we can find a d such that if |y - x| < d, then |f(y) - f(x)| < e. Now, let's fix an x in [a,b] and choose an N such that for all n > N, |f(x) - fn(x)| < e/2. This is possible because fn converges point-wise to f. Next, let's choose a y such that |y - x| < d. Then, we have:

|f(y) - fn(y)| <= |f(y) - f(x)| + |f(x) - fn(x)| < e/2 + e/2 = e

This shows that for all x in [a,b], |f(y) - fn(y)| < e for all n > N, which means that {fn} converges uniformly to f.

Now, let's consider the fact that each fn is nondecreasing. This means that for all x in [a,b], fn(x) <= fn+1(x). In other words, the sequence {fn(x)} is nondecreasing for every x in [a,b]. This is important because it allows us to use