# Prove that there is an inner product on R^2, such that the associated norm is given by: ||(x,y)||=(|x|^p+|y|^p)^(1/p) where p>0 only if p=2

Prove that there is an inner product on ${\mathbb{R}}^{2}$, such that the associated norm is given by:
$\parallel \left(x,y\right)\parallel =\left(|x{|}^{p}+|y{|}^{p}{\right)}^{\frac{1}{p}}$
where p>0 only if p=2So 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:
$\left(|{x}_{1}+{x}_{2}{|}^{p}+|{y}_{1}+{y}_{2}{|}^{p}{\right)}^{\frac{1}{p}}=\left(|{x}_{1}|+|{y}_{1}|{\right)}^{\frac{1}{p}}+\left(|{x}_{2}{|}^{p}+|{y}_{2}{|}^{p}{\right)}^{\frac{1}{p}}$
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 $p\ne 2$).
Since for an inner product, the parallelogram equality must hold,
$⟨u,v⟩=\frac{1}{2}\left(\parallel u+v{\parallel }^{2}+\parallel u-v{\parallel }^{2}\right)$
must also hold.
Using this definition of the inner product, I also tried to show a contradiction by showing that if $p\ne 2$, the Cauchy-Schwarz Inequality didn't hold. However I think that's a dead end.
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Chiecrere2f
Your parallelogram equality does not seem correct. If a norm $‖v‖$ is given by an inner product, then
$‖v+w{‖}^{2}+‖v-w{‖}^{2}=2\left(‖v{‖}^{2}+‖w{‖}^{2}\right)$
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.