Proof involving sequences of functions and uniform convergence

[kidmode01]

Homework Statement

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

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

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

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

Show that:

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

Homework Equations

Theorem:

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

\lim_{n\rightarrow\infty} \int_{E} {f_n} = \lim_{n\rightarrow\infty} \int_{E} F

The Attempt at a Solution

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

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

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

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

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

Looking at our new sequence:

f_n = \phi_n g(x)

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

Thus I think because of the continuity of \phi_n and g, and the uniform convergence of \phi_n:

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

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.

 

[kidmode01]

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:

<br /> \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx = <br /> \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<br />

but

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

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

\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)

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

So I started by breaking my integral up:
<br /> \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<br />

as c goes to zero.

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

<br /> \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)

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)

 

[Dick]

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.

 

[kidmode01]

Right, thanks for pointing that out.

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

<br /> <br /> \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<br /> <br />

and then as c goes to zero:
<br /> 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)

Edit: Ya this is wrong.

I just don't feel right about this though.

 

[kidmode01]

Ah I'm all messed up now lol

 

[Dick]

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'.