How can I show this convergence is uniform?

Question

Let f_n be a sequence of real valued functions that converges on a set X (it can be a subset of R^n, or any metric space) to a function f. Suppose that

(a) The sequence f_n is monotonic (the sequence, not the functions)
(b) X is compact
b) The functions f_n, n=1,2,3... and f are continuous

Show that the convergence of f_n to f is uniform on X.

Thank you for our help

Answer 1

While it's true that you could go look this up, it's actually a fairly simple proof. Here's the outline:

Fix ε > 0; we need to find an N such that for all x, |fn(x) - f(x)| < ε for n > N.
For each x, let N(x) be the N given by (ordinary) convergence corresponding to ε/2.

The essential step: For each x in X, find neighborhoods B(x) of x where |fn(y) - f(y)| < ε for all n > N(x), by monotonicity and continuity.

Now, just find a finite collection of the B(x)'s that cover X by compactness. Let δ = min {diam(B(x))} over the finite set, and choose x* such that diam(B(x*)) = δ. Then N(x*) is your desired N.
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Answer 2

f_n(x) isn't differentiable at 0 for any value of n. to confirm that it to be differentiable at a factor, it's going to be non-supply up at that factor, yet no f_n(x) is even defined at 0, considering that f_n(0) = (-a million/n)^a million/2. till we are talking approximately complicated diagnosis right here, you are able to't take the sq. root of a unfavorable.

Answer 3

This is Dini's theorem, a classical result. You find the proof in analysis books.