# Re: equicontinuous proof check,

• From: José Carlos Santos <jcsantos@xxxxxxxx>
• Date: Mon, 09 Apr 2007 09:06:37 +0100

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
.