# Thread: Why does this limit of this integral work?

1. ## Why does this limit of this integral work?

"If we set epsilon > 0 and N as a large natural number, and if j, k > N:

$\displaystyle \left| \int_{a}^{b} f_j(x) \, dx \right| \, -\, \left| \int_{a}^{b} f_k(x) \, dx \right| \, \leq \int_{a}^{b} \left| f_j(x) - f_k(x) \right| dx$

OK where exactly is this coming from? This appears as one of the initial statements in a proof about f_j's integral approaching f's integral. But I could not find the theorem that states this. Is it supposed to be self-evident or something?

2. ## Re: Why does this limit of this integral work?

all you've written here is the triangle inequality where $\left|\displaystyle{\int_a^b}g(t)~dt\right|$ is a norm of $g(t)$

3. ## Re: Why does this limit of this integral work?

Originally Posted by romsek
all you've written here is the triangle inequality where $\left|\displaystyle{\int_a^b}g(t)~dt\right|$ is a norm of $g(t)$
Ah, OK. The triangle inequality for integrals. Got it.

4. ## Re: Why does this limit of this integral work?

It's a little more complicated than that. Using the form of the triangle inequality $\displaystyle |y|-|z| \leq |y-z|$,

$\displaystyle \left| \int_{a}^{b} f_j(x) \, dx \right| \, -\, \left| \int_{a}^{b} f_k(x) \, dx \right| \, \leq \left| \int_{a}^{b} f_j(x) \, dx - \int_{a}^{b} f_k(x) \, dx \right|= \left| \int_a^b f_j(x)-f_k(x) \,dx \right|$

And using $\displaystyle \left| \int a \right| \leq \int |a|$ gives

$\displaystyle \left| \int_{a}^{b} f_j(x) \, dx \right| \, -\, \left| \int_{a}^{b} f_k(x) \, dx \right| \leq \int_a^b |f_j(x)-f_k(x)| \,dx$

- Hollywood