How to interpret the integral of the absolute value?

[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:

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

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

\begin{equation}
||x_n||^2 = \int_0^1 1-2nt-n^2t^2 dt = t - nt^2 -n^2t^2/3 = 1 - n - n^2/3
\end{equation}

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

Thanks!

 

[Buffu]

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

 

[SeM]

Ok! Thanks! Then its a typo.

 

[Buffu]

That's what I thought, then either the book is wrong (which I doubt), or this is a different notation. Let me scan it and upload it here in the next reply.

Wait, how did you get books answer ?

 

[SeM]

Wait, how did you books answer ?

Attached

 

[fresh_42]

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:

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

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

\begin{equation}
||x_n||^2 = \int_0^1 1-2nt-n^2t^2 dt = t - nt^2 -n^2t^2/3 = 1 - n - n^2/3
\end{equation}

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

Thanks!

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}}##.

 

[Buffu]

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}}##.

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

 

[SeM]

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

 

[SeM]

Wait, how did you get books answer ?

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?

 

[Buffu]

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?

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

 

[SeM]

Ah! Indeed! Thanks!