Uniform Convergence, help with proof of theorems

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Homework Statement


1.) Prove that if { [tex]f_{n}[/tex] } is a sequence of functions defined on a set D, and if there is a sequence of numbers [tex]b_{n}[/tex], such that [tex]b_{n}[/tex] [tex]\rightarrow[/tex] 0, and | [tex]f_{n}[/tex](x) | [tex]\leq[/tex] [tex]b_{n}[/tex] for all x [tex]\in[/tex] D, then { [tex]f_{n}[/tex] } converges uniformly to 0 on D.

2.) Prove that if { [tex]f_{n}[/tex] } is a sequence of functions defined on a set D, and if { [tex]f_{n}[/tex] } converges uniformly to zero on D, then { [tex]f_{n}[/tex]([tex]x_{n}[/tex]) } converges to zero for every sequence { [tex]x_{n}[/tex] } of points of D.


Homework Equations



Definition of Uniform Convergences:

{ [tex]f_{n}[/tex] } is said to converge uniformly on D to a function f if, for each [tex]\epsilon[/tex] > 0, there is N such that,

| f(x) - [tex]f_{n}[/tex](x) | < [tex]\epsilon[/tex] whenever x [tex]\in[/tex] D and n>N.



The Attempt at a Solution



1.) SO for this one, it seems that i can just pick N = [tex]b_{n}[/tex] = [tex]\epsilon[/tex] and then this theorem follows immediately from the definition of uniform convergence.

2.) Similarly, for this one, because [tex]x_{n}[/tex] is just a sequence of points in D, then we just replace x with [tex]x_{n}[/tex] and the proof will also follow immediately from the definition of uniform convergence.

Am I on the right track with these? They both seem some what trivial.
 

Answers and Replies

  • #2
jbunniii
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1.) SO for this one, it seems that i can just pick N = [tex]b_{n}[/tex] = [tex]\epsilon[/tex] and then this theorem follows immediately from the definition of uniform convergence.

Not exactly. The [itex]\epsilon[/itex] isn't necessarily equal to any of the [itex]b_n[/itex] (and certainly [itex]N[/itex] isn't equal to either of them!), but can be any positive real number. But then you know that you can find an [itex]N[/itex] such that [itex]b_n \leq \epsilon[/itex] for all [itex]n \geq N[/itex], and the result follows from this.

2.) Similarly, for this one, because [tex]x_{n}[/tex] is just a sequence of points in D, then we just replace x with [tex]x_{n}[/tex] and the proof will also follow immediately from the definition of uniform convergence.

Pretty much, although I can't say for sure unless you write down exactly what your argument is.
 
  • #3
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For #1, we know that bn converges to zero, and thus by definition of convergence, i can pick N = eps. The fact that bn is only convergent and not uniformly convergent is confusing me as to why fn(x) being less than or equal to bn implies UNIFORM convergence.
 
  • #4
jbunniii
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For #1, we know that bn converges to zero, and thus by definition of convergence, i can pick N = eps. The fact that bn is only convergent and not uniformly convergent is confusing me as to why fn(x) being less than or equal to bn implies UNIFORM convergence.

I'm not sure what you mean by [itex]N = \epsilon[/tex]. If you picture plotting the sequence [itex]\{b_n\}[/itex] on a graph, with [itex]n[/itex] along the horizontal axis, and the value of [itex]b_n[/itex] on the vertical axis, then [itex]N[/itex] is a horizontal-axis value, whereas [itex]\epsilon[/itex] is a vertical-axis value. It doesn't make any sense to set one equal to the other.

As to why [itex]f_n(x) < b_n[/itex] implies uniform convergence, it is because [itex]b_n[/itex] is a constant that does not depend on [itex]x[/itex].
 
  • #5
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ok, so i figured out #1, I used the fact that |fn(x)| is less than or equal to bn if and only if it is greater than or equal to -bn, and less than or equal to bn, then applied the squeeze theorem. Still working on #2
 

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