Norm from a scalar product only for $p = 2$

Question

How to prove that the following norm, defined on $C([0,1],\mathbb{C})$ :

$||f||_p = (\int_0^1 |f(t)|^p dt)^{1/p}$ (for $p \in [1,+\infty[$) is "from" a scalar product only if $p=2$ ?

I don't know how to show it at all, I need food for thought... Thank you in advance !

Two questions, as food for thought: (1) can you define "scalar product"? (2) can you write down precisely what you mean by "from"? (In other words, can you write down a formula for what the scalar product $\langle f,g\rangle$ will be if it exists?) — Billy, Mar 29 '18 at 14:09

score 3 · Accepted Answer · answered Mar 29 '18 at 14:09

3

Since your name is French, I assume with scalar product (produit scalaire) that you're talking about the inner product.

You could use that the parallelogram law holds if and only if the norm is induced by an inner product. So you only need to check that $$2\|f\|_p^2+2\|g\|_p^2=\|f+g\|_p^2+\|f-g\|_p^2 $$ holds for all $f,g\in C([0,1],\mathbb{C})$ only for $p=2$.

This post contains more details about the proof of the parallelogram law.

answered Mar 29 '18 at 14:09

The Phenotype

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1

I really thank you ! Indeed, "scalar product" is "inner product" and "from" means "induced by" in my post. I'll see the topic you link :) – Mélanie De la Cheminée Mar 29 '18 at 14:30

Norm from a scalar product only for $p = 2$

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