Re: uniformly equicontinuous
- From: "Robert Israel" <israel@xxxxxxxxxxx>
- Date: 14 Jun 2006 22:30:25 -0700
NotP wrote:
"NotP" <spam@xxxxxxxx> wrote in message
news:fz5kg.38504$8G3.26148@xxxxxxxxxxxxxxxxxxxxxxx
A family of functions F is uniformly equicontinuous if for all e>0 there
exists d>0 such that for all f in F, d(x,y)<d implies d(f(x),f(y))<e.
Suppose g: Rx[0,1] -> R is a bounded, continuous function. Is the family
of functions
{f_y : R -> R | y in [0,1]} , where f_y(x) = g(x,y), uniformly
equicontinuous?
If so, suggestions for a proof? If not, counterexample?
Thanks.
Oops. I forgot an extra condition.
For each y in [0,1], g(x,y) -> 0 as x-> infinity, and as
x-> -infinity.
Hint: try for a counterexample where g(x,y) = 1 on a certain curve,
while
g(x,y) = 0 outside a certain region containing the curve.
Robert Israel
israel@xxxxxxxxxxxxxxxxx'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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