# The only norm induced by inner product vector space is 2-norm. Why?

I was looking at a Linear Algebra textbook and I encountered this question. I tried to prove it, but I didn't succeed. The question is:

The Hölder inequality states that for any $$p, q \in[1, \infty)$$ with $$\frac{1}{p}+\frac{1}{q}=1$$ and any $$\psi=\left(a_{1}, \ldots, a_{n}\right)^{T} \in \mathbb{C}^{n}$$ and $$\phi=\left(b_{1}, \ldots, b_{n}\right)^{T} \in \mathbb{C}^{n},$$So we have: $$\begin{array}{l} \\ \qquad \sum_{x=1}^{n}\left|a_{x} b_{x}\right| \leqslant\|\psi\|_{p}\|\phi\|_{q} \end{array}$$ Use this to show that $$\|\cdot\|_{p}$$ is a norm, and that for $$p \neq 2$$ this norm is not induced from an inner product.

• If you have an inner product you can get an orthonormal basis. In terms of this basis the norm has the usual form, which is the form for p= 2. Sep 27, 2020 at 15:58

$$\|x\|^2+\|y\|^2=\frac{1}{2}((\|x+y\|^2+\|x-y\|^2)$$
It is not difficult to show that this does not hold for the $$p$$-norm when $$p\neq 2$$.