# Thread: Estimate an integral

1. ## Estimate an integral

Hello everyone.

It is $a,b,q \in \mathbb{R}, a and $q \ge 1$

Furthermore let $c \in ]a,b[ \mbox{ and } j_0 \in \mathbb{N} : c-(1/j_0) > a$.

$f_j:[a,b]\to [0,1]$ is for every $j \ge j_0$ a continious function, e.g.

$f_j=\begin{cases} 0, & \mbox{iff } a\le x \le c-\frac{1}{j+1} \\ 1, & \mbox{iff } c

for all $i \ge j \ge j_0$ it is

$||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q$

I don't get the following lines:

$= \int^{\displaystyle c-\frac{1}{i+1}}_{\displaystyle c-\frac{1}{j+1}} \ f_j^q+\int^c_{\displaystyle c-\frac{1}{i+1}}\ (f_j-f_i)^q$

I don't unterstand the next line either:

$\le \frac{1}{j+1}+\frac{2^q}{i+1}$

Can anyone explain it to me, please?

Kind regards
Rapha

2. You have a definition, I assume, of a norm
$

||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q
$

that has rules on how to evaluate it -
f = 0 for x between a and c-...
1 for x between, and so on.

Now the part you didn't understand is trying to do the integral to evaluate the norm. Note the norm integral is from a to b. For x just a little larger than a the f's are zero, so there is no need to do an integral. For x just a little less than c, the f is the really complicated expression, so the integral is just written in terms of the f's. For x in the little space between 1/j+1 and 1/i+1, one of the f's is zero so you can ignore it.

I suspect the next line is actually doing the integral.

3. Originally Posted by qmech
You have a definition, I assume, of a norm
$

||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q
$

that has rules on how to evaluate it -
f = 0 for x between a and c-...
1 for x between, and so on.

Now the part you didn't understand is trying to do the integral to evaluate the norm. Note the norm integral is from a to b. For x just a little larger than a the f's are zero, so there is no need to do an integral. For x just a little less than c, the f is the really complicated expression, so the integral is just written in terms of the f's. For x in the little space between 1/j+1 and 1/i+1, one of the f's is zero so you can ignore it.

I suspect the next line is actually doing the integral.
That makes sense.
Thank you very much, qmech!