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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.