hilbert space

**surjective** · March 15th 2011, 01:35 PM

Hello,

I am considering the banach-space $L^{p}(\mathbb{R})$ and the functions:

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

I have calculated the numbers $\Vert f \Vert_{p}$ , $\Vert g \Vert_{p}$ , $\Vert f+g \Vert_{p}$ and $\Vert f-g \Vert_{p}$ (they all have value 1).

How do I, using the above numbers, conclude that $L^{p}(\mathbb{R})$ does not form a hilbertspace for $p \neq 2$ ?

Thanks.

**Opalg** · March 15th 2011, 01:58 PM

The parallelogram law holds in a Hilbert space.

And BTW they shouldn't all have the value 1.

**surjective** · March 15th 2011, 02:11 PM

Ok thanks. Maybe I made a mistake in calculating the values. Let me look at it.

**surjective** · March 15th 2011, 02:37 PM

Here are my values:

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

$\Vert g \Vert_{p}=\left( \int_{1}^{2}|1|^{p}dx \right)^{\frac{1}{p}}=1$

$\Vert f+g \Vert_{p}=\left( \int_{0}^{1}|1|^{p}dx + \int_{1}^{2}|1|^{p}dx\right)^{\frac{1}{p}}=2$

Could you check the last calculation???

**Opalg** · March 16th 2011, 12:58 AM

Originally Posted by surjective

Here are my values:

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

$\Vert g \Vert_{p}=\left( \int_{1}^{2}|1|^{p}dx \right)^{\frac{1}{p}}=1$

$\Vert f+g \Vert_{p}=\left( \int_{0}^{1}|1|^{p}dx + \int_{1}^{2}|1|^{p}dx\right)^{\frac{1}{p}}=2$

Could you check the last calculation???

The first two are correct. For the last one, the sum of the two integrals inside the parentheses is 2, but that should then be raised to the power 1/p. So the answer should be $2^{1/p}.$

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