# Why does this limit of this integral work?

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• March 16th 2014, 06:43 PM
phys251
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:

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

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? :confused:
• March 16th 2014, 07:13 PM
romsek
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 $g(t)$
• March 16th 2014, 08:08 PM
phys251
Re: Why does this limit of this integral work?
Quote:

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.
• March 17th 2014, 07:46 AM
hollywood
Re: Why does this limit of this integral work?
It's a little more complicated than that. Using the form of the triangle inequality $|y|-|z| \leq |y-z|$,

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

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

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

- Hollywood