Re: equicontinuous proof check,
 From: José Carlos Santos <jcsantos@xxxxxxxx>
 Date: Mon, 09 Apr 2007 09:06:37 +0100
On 09042007 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 xy < 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
.
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