00:01
Okay, so let's get started by showing that fn, because she implies that there exists a function f such that fn converges uniformly to f.
00:18
Okay, how can we do this? well, this is pretty easy.
00:23
Let's observe that limit as n goes to infinity of fn of x exists for every x belonging to d.
00:39
Because the sequence fn of x okay let me write because the sequence of real numbers fn of x is cauchy and we know that every cauchy sequence has a limit okay now at this point point we can define f of x as limit as n goes to infinity of fn of x.
01:18
Well we need to show that fn converges uniformly to f.
01:24
We already know that fn converges pointwise to f by construction action so we just need to show that this convergence is uniform.
01:36
Well to do this let's keep in mind that for every epsilon greater than zero there exists capital n positive integer such that that the supremum over d of fn of x minus fm of x is less than epsilon for every n and m greater than or equal to capital n.
02:09
Perfect.
02:10
Okay, here let me write the supremum with with x belonging to d.
02:18
Okay, well this inequality here shows that the supremum with x belonging to d of fn of x minus limit as m goes to infinity of fm of x is less than epsilon for every n greater than or equal to capital n but well by construction this guy here is f of x so we have shown that for every epsilon greater than zero there exists capital n positive integer such that the supremum with x belonging to d of the the absolute value of fn of x minus f of x is less than epsilon.
03:21
Perfect.
03:22
This is just the definition of uniform convergence.
03:27
So, fn converges uniformly to f...