# Banach-spaces

• March 23rd 2010, 12:35 PM
surjective
Banach-spaces
Hello,

I have a series of questions. Some of them may seem easy but I am having trouble understanding them. Here is the first:

I am considering the Banach-space $L^p(\mathbb{R})$ for some $p\geq 1$ and the functions $f$ and $g$:

$f=\chi_{[0,1[}$
$g=\chi_{[1,2[}$

I know that the characteristic function $\chi_{E}(x)$ for the set $E$ is given as:

$\chi_{E}(x)=\left\lbrace \begin{matrix} 1, x \in E \\ 0, x \ni E \end{matrix} \right.$

I also know that the norm on $L^p(\mathbb{R})$ is given by:

$\Vert f \Vert_{p}$= $\left( \int_{-\infty}^{\infty}|f(x)|^{p}dx \right)^{\frac{1}{p}}$

Could someone kindly explain to me how to compute $\Vert f \Vert_{p}$ and $\Vert f+g \Vert_{p}$ using the functions defined above.

Thanks alot.
• March 23rd 2010, 02:10 PM
mabruka
It is straightforward,

$\Vert f \Vert_{p}=\Vert\chi_{[0,1[}\Vert_{p}$
$=\left( \int_{(-\infty,\infty)}|\chi_{[0,1[}|^{p}dx \right)^{\frac{1}{p}}$

$=\left( \int_{[0,1]}1^{p}dx \right)^{\frac{1}{p}}$

$=\left(1-0 \right)^{\frac{1}{p}}=1$

What i have done is give two disjoint sets whose union is the total,

$\mathbb R = [0,1] \cup [0,1]^c$

f is constant on each of these two sets. 1 in the first and 0 in the second.

Integrating the constant 1 over any set yields the measure of that set, in your problem that is the lebesgue measure, and since our set is an interval, the lebesgue measure of an interval is the length of such interval. Hence 1 - 0 .

Although the other set has positive lebesgue measure, the fact that f is 0 over it vanishes that part.

Do the same with the other and post your results.
• March 23rd 2010, 06:13 PM
surjective
Banach-space
Hey,

Here is what I have done with $f+g$:

$\left( \int_{]-\infty,\infty[} \vert\chi_{[0,1[}+\chi_{[1,2[} \vert^{p} dx \right)^{\frac{1}{p}}$ =

$\left( \int_{[0,1]} \vert\chi_{[0,1[}+\chi_{[1,2[} \vert^{p} dx \right)^{\frac{1}{p}}$ + $\left( \int_{[0,1]^{c}} \vert\chi_{[0,1]^{c}}+\chi_{[1,2]^{c}} \vert^{p} dx \right)^{\frac{1}{p}}$ =

$\left( \int_{[0,1]} \vert 1+1 \vert^{p} dx \right)^{\frac{1}{p}}$ + $\left( \int_{[0,1]^{c}} \vert 0+0 \vert^{p} dx \right)^{\frac{1}{p}}$ =

$\left( \int_{0}^{1} 2^{p} dx \right)^{\frac{1}{p}}$ =

$\left( [2^{p}x]_{0}^{1} \right)^{\frac{1}{p}}$ =

2

What do you think ?

So far I have that:

$\Vert f \Vert_{p}$ = 1

$\Vert g \Vert_{p}$ = 1

$\Vert f+g \Vert_{p}$ = 2

$\Vert f-g \Vert_{p}$ = 0

My next question is: How do I show that $L^{p}(\mathbb{R})$ does not form a Hilbert-space for $p\neq 2$.

$\text{Thanks}^{\infty}$
• March 24th 2010, 05:42 AM
HallsofIvy
Quote:

Originally Posted by surjective
Hey,

Here is what I have done with $f+g$:

$\left( \int_{]-\infty,\infty[} \vert\chi_{[0,1[}+\chi_{[1,2[} \vert^{p} dx \right)^{\frac{1}{p}}$ =

$\left( \int_{[0,1]} \vert\chi_{[0,1[}+\chi_{[1,2[} \vert^{p} dx \right)^{\frac{1}{p}}$ + $\left( \int_{[0,1]^{c}} \vert\chi_{[0,1]^{c}}+\chi_{[1,2]^{c}} \vert^{p} dx \right)^{\frac{1}{p}}$ =

$\left( \int_{[0,1]} \vert 1+1 \vert^{p} dx \right)^{\frac{1}{p}}$ + $\left( \int_{[0,1]^{c}} \vert 0+0 \vert^{p} dx \right)^{\frac{1}{p}}$ =

No! The two functions are NOT equal to 1 on [0, 1]. $\chi_{[0,1[}= 1$ on [0,1] but $\chi_{[1,2[}= 0$ on [0,1].

You need three integrals- on [0,1], on [1, 2] and on the complement of the union of those two intervals.

Quote:

$\left( \int_{0}^{1} 2^{p} dx \right)^{\frac{1}{p}}$ =

$\left( [2^{p}x]_{0}^{1} \right)^{\frac{1}{p}}$ =

2

What do you think ?

So far I have that:

$\Vert f \Vert_{p}$ = 1

$\Vert g \Vert_{p}$ = 1

$\Vert f+g \Vert_{p}$ = 2

$\Vert f-g \Vert_{p}$ = 0

My next question is: How do I show that $L^{p}(\mathbb{R})$ does not form a Hilbert-space for $p\neq 2$.

$\text{Thanks}^{\infty}$
• March 24th 2010, 02:45 PM
mabruka
You must use the next partition of R

(-\infty,0), [0,1), {1}, (1,2], (2,\infty)

And use the values of f and g on each one of the above sets:

f g
(-\infty,0) 0 0
[0,1) 1 0
{1} 1 1
(1,2] 0 1
(2,\infty) 0 0

so integration over R means integrating over each set where f and g take constant values.

Dont forget that while integratin over {1} , f and g take constant value 1, the set itself has lebesgue measure 0 so integrating anything over it gives 0.