# Definition of norm when dealing with inner product space [duplicate]

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Is it possible to have a norm in an inner product space, such that $\lVert v\rVert\neq\sqrt{\langle v,v\rangle}$?

In my linear algebra class, we've defined $\lVert v\rVert$ to be equal to $\sqrt{\langle v,v\rangle}$, for each $v\in V$, where we were dealing with an inner product space $(V,\langle.,.\rangle)$.

However, now in a more advanced class in Analysis where we're going to work in multiple dimensions, the norm is defined on a vector space, such that certain properties hold. Does this mean that the definition in my Linear Algebra book was too specific, in the sense that it's also possible to have a norm in an inner product space where the above equality doesn't hold? Or is it true that for each inner product space, the norm equals $\sqrt{\langle v,v\rangle}$?

## marked as duplicate by Watson, Community♦Feb 8 '17 at 12:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Take $\ell^3(\Bbb N)$ or $\Bbb R^n$ with a $p$-norm with $p \neq 2$. – Watson Feb 8 '17 at 12:24
• – jibounet Feb 8 '17 at 12:27

## 1 Answer

If a norm is defined by an inner product the way you write in the question, the parallelogram law holds. It's possible to check that e.g. only $p$-norm on $\mathbb{R}^d$:$$\left\| x \right\|_p = \left( \sum \limits_{i=1}^n \left| x_i \right|^p \right)^{1/p}$$ that satisfies this is $p=2$.