Re: equicontinuous proof check,



On 09-04-2007 6:55, vsgdp wrote:

If (f_n) is a uniformly equicontinuous sequence of functions on a
compact interval and (f_n) converges to f pointwise, prove (f_n)
converges to f uniformly.

equicontinuous means

You should have written that this is what *uniformly* equicontinuous
means.

for all e>0 there exists a d such that |x-y| < d implies |f_n(x) -
f_n(y)|<e for all x,y,n

|f_n(x) - f(x)| <= |f_n(x) - f_n(y)| + |f_n(y) - f(x)|

We can make |f_n(x) - f_n(y)| < e/2 from equicontinuous

|f_n(y) - f(x)| <= |f_n(y) - f(y)| + |f(y) - f(x)| < e/4 + e/4 = e/2

f_n(y)-->f(y) pointwise and uniform continuity respectively.

There are two problems here. First of all, it is true that

|f_n(y) - f(y)| < e/4

if _n_ is large enough. To be more precise, there is some _p_ such
that that inequality holds if n >= p. The problem here is that _p_
depends upon _y_.

The second problem is this: how do you prove that

|f(y) - f(x)| < e/4?

Your argument is "by uniform continuity". Uniform continuity of what?
Of _f_? You didn't even prove that it is continuous!

Best regards,

Jose Carlos Santos
.



Relevant Pages

  • Re: uniformly bounded function sequenc...
    ... and it converges pointwise to a continuous function f, ... Best regards, ... Jose Carlos Santos ...
    (sci.math)
  • Re: Could someone translate this to Plain English?
    ... What means "pointwise fixed"? ... (and as Knuth wrote at the passage that you quoted), ... Best regards, ... Jose Carlos Santos ...
    (sci.math)
  • Re: equicontinuous proof check,
    ... Your argument is "by uniform continuity". ... And it is uniform continuity since it is on a compact set. ... and some sequence _k of points of your interval ... Jose Carlos Santos ...
    (sci.math)
  • Re: equicontinuous proof check,
    ... I don't see how to fix that problem. ... Your argument is "by uniform continuity". ... And it is uniform continuity since it is on a compact set. ... Jose Carlos Santos ...
    (sci.math)