# B How to interpret the integral of the absolute value?

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1. Dec 21, 2017

### SeM

This is rather basic, and may be a misconception of the notation, however, I can't make the following sum up:

The following is given:

x_n(t) = 1 -nt , (if 0 <= t <= 1/n) and 0, (if 1/n < t <= 1)

However, this part I can't grasp this part in the book:

||x_n||^2 = \int_0^1 |x_n(t)|^2 dt = \frac{1}{3n}

I tried it, and got a different answer, where i integrated $|x_n(t)|^2=(1-nt)^2 = 1-2nt-n^2t^2$:

||x_n||^2 = \int_0^1 1-2nt-n^2t^2 dt = t - nt^2 -n^2t^2/3 = 1 - n - n^2/3

The right answer is however given in the first integral. What did I do wrong here?

Thanks!

2. Dec 21, 2017

### Buffu

It should be $$||x_n||^2 = \int_0^{1/n} 1-2nt+n^2t^2 dt$$

Last edited: Dec 21, 2017
3. Dec 21, 2017

### SeM

Ok! Thanks!! Then its a typo.

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4. Dec 21, 2017

### Buffu

Wait, how did you get books answer ?

5. Dec 21, 2017

### SeM

Attached

6. Dec 21, 2017

### Staff: Mentor

There is a sign error, since $(1-nt)^2=1-2nt+n^2t^2$ and as @Buffu said: $||x_n||^2 = \int_0^1 = \int_0^{\frac{1}{n}}$.

7. Dec 21, 2017

### Buffu

Yes, I also copied the error. Now corrected it in my post.

8. Dec 21, 2017

### SeM

Thanks! I have corrected it in the book. It's Kreyszig Introduction to Functional Analysis, as part of the discussions with George Jones.

Last edited by a moderator: Dec 21, 2017
9. Dec 21, 2017

### SeM

Hi Buffu, something is odd with the step "and the quotient" yielding the answer >n. Here I got: n/1/3n = 3n^2. but the book says differently again. It appears as an error, can you see if this is yet another error?

10. Dec 21, 2017

### Buffu

You forgot to take the square root. The integral gives norm squared; you need ratio of norms not of norms squared.

Last edited: Dec 21, 2017
11. Dec 21, 2017

### SeM

Ah! Indeed! Thanks!