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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.

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  • $\begingroup$ 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. $\endgroup$
    – Peter
    Sep 27, 2020 at 15:58

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A norm that is induced by an inner product satisfies the parallelogram identity, which for spaces over the reals has the form:

$$\|x\|^2+\|y\|^2=\frac{1}{2}((\|x+y\|^2+\|x-y\|^2)$$

It is not difficult to show that this does not hold for the $p$-norm when $p\neq 2$.

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