Proof involving sequences of functions and uniform convergence

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Homework Statement


Let [tex] \phi_n(x) [/tex] be positive valued and continuous for all x in [tex] [-1,1] [/tex] with:

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = 1 [/tex]

Suppose further that [tex] \{\phi_n(x)\} [/tex] converges to 0 uniformly on the intervals [tex] [-1,-c] [/tex] and [tex] [-c,1] [/tex] for any [tex] c > 0 [/tex]

Let g be any function which is continuous on [tex] [-1,1] [/tex].

Show that:

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx = g(0) [/tex]


Homework Equations



Theorem:

If the functions [tex] \{f_n\} [/tex] and F are integrable on a bounded closed set E, and [tex] \{f_n\} [/tex] converges to F pointwise on E, and if [tex] \Vert {f_n} \Vert < M [/tex] for some M and all n = 1,2,3,..., then:

[tex] \lim_{n\rightarrow\infty} \int_{E} {f_n} = \lim_{n\rightarrow\infty} \int_{E} F [/tex]



The Attempt at a Solution



Alrighty, so the first thing I did was apply the theorem on:

[tex] f_n = \phi_n [/tex] on the intervals [-1,-c] and [-c,1]

and came up with the following integrals(since uniform convergence implies pointwise convergence):

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) = 0 [/tex]
and
[tex] \lim_{n\rightarrow\infty} \int_{-c}^{-1} \phi_n(x) = 0 [/tex]

If we add these integrals together we know what happens at c=0.

Looking at our new sequence:

[tex] f_n = \phi_n g(x) [/tex]

This sequence is uniformly bounded on [-1,-c] and [-c,1] and converges pointwise to 0.

Thus I think because of the continuity of [tex] \phi_n [/tex] and g, and the uniform convergence of [tex] \phi_n [/tex]:

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) = 0 [/tex]
[tex] \lim_{n\rightarrow\infty} \int_{-c}^{-1} \phi_n(x) g(x) = 0 [/tex]

Again if we add these integrals together we know what happens when c = 0. But I'm not entirely sure where to go from here or if I'm going in the right direction. Any help is appreciated.
 

Answers and Replies

  • #2
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Second attempt at solution

Okay so I've tried a little bit more now:

Just adding and subtracting a term from the limit we want to prove:

[tex]
\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx =
\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx + \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx
[/tex]

but

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx = g(0) [/tex]

and by the mean value theorem for integrals we know there exists a t in [tex] [-1,1] [/tex] such that:

[tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx = (g(t)-g(0))\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = g(t) - g(0) [/tex]

and now I want to make an argument that states g(t) must be g(0).

So I started by breaking my integral up:
[tex]
\lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{-c}^{1} \phi_n(x) g(x) = 0
[/tex]

as c goes to zero.

But then g(t) - g(0) = 0 and thus g(t) = g(0) and then we would have:

[tex]
\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx = \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx + \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx = g(t) - g(0) + g(0) = g(0) [/tex]

The very last step seems a little hand wavy, I hope someone can help me.

Edit: (I also realize some of the stuff in my first post is wrong, but it won't let me edit it)
 
Last edited:
  • #3
Dick
Science Advisor
Homework Helper
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You MUST mean phi_n converges uniformly to zero on [-1,-c] and [c,1] for any c>0. Not [-1,-c] and [-c,1]. If that were the case then the integral of phi_n would go to zero as n->infinity.
 
  • #4
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Right, thanks for pointing that out.

Then my integrals at the bottom of my second post change to:

[tex]

\lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{c}^{1} \phi_n(x) g(x) = 0

[/tex]

and then as c goes to zero:
[tex]
0 = \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{c}^{1} \phi_n(x) g(x) = \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx = (g(t)-g(0))\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = g(t) - g(0) [/tex]

Edit: Ya this is wrong.

I just don't feel right about this though.
 
Last edited:
  • #5
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Ah I'm all messed up now lol
 
  • #6
Dick
Science Advisor
Homework Helper
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g(x) is continuous at 0. So if c is small enough g(x) is 'almost' g(0) on [-c,c]. phi_n is 'almost' zero outside of [-c,c] for large enough n. Now put in some epsilons to quantify 'almost'.
 

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