uniform convergence?

Question

Let f, f1, f2, ... be continuous real valued functions on the compact metric space E with lim fn = f.
prove that if f1(p)<= f2(p)<= ... for all p in E
the the sequence f1,f2,f3,... converges uniformly

Answer 1

It is fun to notice that the result is false if any of the hypotheses are left out. For example, here is a counterexample for when the limit function is not continuous. (The sketch of the proof of the result you want is below. You should be able to fill in the details if you are at a level to ask this question.)

For the space E, take the closed interval [0,1]. For f1, take the piecewise defined function that is equal to the line y=2x-1 on [0,1/2] and is equal to 0 on [1/2,1]. Generally, take fn to be the piecewise defined function equal to the line y=2^n x-1 on [0,1/2^n] and is equal to 0 on [1/2^n,1]. These functions satisfy your conditions and converges pointwise to the function that is equal to -1 for x=0 and 0 for every x in (0,1]. But it is clear that the convergence is not uniform.

To prove this correct result, first simplify to the case that the functions decrease pointwise to zero. (If the limit function is continuous, then fn-f is also continuous, so replace fn by fn-f to get a sequence of continuous functions going pointwise to 0. The decreasing part is ok too; since fn-f is increasing as a sequence, -(fn-f) is decreasing. So replace fn by -(fn-f) and now you have a sequence of functions decreasing pointwise to 0.) Then let epsilon be positive and use the definitions of pointwise convergence to find an N_x for every x in your space, use the definition of continuity of fN_x to get a delta_x for every point, use the definition of compact to cover your space by a finite number of the delta_x-balls (balls of radius delta_x centered at x), and then use the monotonicity of the sequence of functions, recalling that all of your function values are necessarily positive, by choice of "decreasing to zero". All of the hypotheses are crucial...

Answer 2

start off at the bottom of this
http://www.mathreference.com/lc-ser,unic.html

and step back until you have the full chain.

Best of luck - Mike

Answer 3

huh ? this is by definition.