# Proof involving sequences of functions and uniform convergence

## 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.

## Answers and Replies

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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:

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

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

as c goes to zero.

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

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

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)

Last edited:
Dick
Science Advisor
Homework Helper
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.

Right, thanks for pointing that out.

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

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

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

Edit: Ya this is wrong.

I just don't feel right about this though.

Last edited:
Ah I'm all messed up now lol

Dick
Science Advisor
Homework Helper
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'.