Prove that there is an inner product on $\mathbb{R}^2$, such that the associated norm is given by:

$ \parallel (x,y) \parallel = (|x|^p + |y|^p)^\frac{1}{p}$

where $ p > 0 $ only if $ p = 2 $

So far what I have tried to do is assume there exists such an inner product for some aribtrary $p$, and then show that some property that holds for all inner products (e.g. the Cauchy-Schwarz inequality, triangle inequality, parallelogram equality, does not hold for the inner product with the said associated norm unless $ p = 2 $.

First I tried the parallelogram equality and ended up with:

$ (|x_1 + x_2|^p + |y_1 + y_2|^p)^\frac{1}{p} = (|x_1| + |y_1|)^\frac{1}{p} +(|x_2|^p + |y_2|^p)^\frac{1}{p}$

but I don't know how to show that this equality only holds for $ p = 2 $ (although I'm pretty sure it does because I tried plugging in random values for $p \neq 2 $ ).

Since for an inner product, the parallelogram equality must hold,

$ \langle u,v \rangle = \frac{1}{2} (\parallel u + v \parallel ^ 2 + \parallel u - v \parallel ^ 2) $

must also hold.

Using this definition of the inner product, I also tried to show a contradiction by showing that if $ p \neq 2 $, the Cauchy-Schwarz Inequality didn't hold. However I think that's a dead end.

  • $\begingroup$ If you can find just one pair of vectors for which the parallelogram equality does not hold for any $p\ne 2$, you are done. $\endgroup$ Jun 19 '15 at 19:20
  • $\begingroup$ I'm not sure I see how that's true? If I find a pair of vectors for $ p = 1 $, that says nothing about $ p = 3 $, unless I'm missing something? $\endgroup$
    – imu96
    Jun 19 '15 at 19:26

Your parallelogram equality does not seem correct. If a norm $\| v \|$ is given by an inner product, then $$ \|v+w\|^2 + \|v-w\|^2 = 2 (\|v\|^2 + \|w\|^2) $$ Now just pick $v = (1,0)$, $w=(0,1)$, and calculate both sides with arbitrary $p$, then you will get an equation which holds iff $p=2$.


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