Is it true that for an inner product space a norm of a vector is defined unambiguously?

Question

Suppose we have some vector space $V$ for which we have defined an inner product $\langle \cdot\rangle$. Thus we have an inner product space. Is it true that $\forall x \in V : \lVert x\rVert := \sqrt{\langle x,x\rangle}$ is the only possible norm in $V$?

Let's consider the following example, that shows, what happens if it is not true (if by an inner product induced norm is not the only possible in this vector space):

Let $\mathcal{P}$ be a vector space of polynomials with an inner product defined as $$\langle P,Q\rangle=\int_0^1P(x)Q(x)\operatorname{d}x \quad \forall P,Q\in\mathcal{P}$$ Suppose we want to find an angle $\phi$ between two polynomials $P_1$ and $P_2$. We use a standard formulae for an angle between two vectors $$\cos\phi={\langle P_1,P_2\rangle \over \rVert P_1\lVert\rVert P_2\lVert}$$ So the norms in denominator should be the ones induced by the inner product, otherwise the angle will depend on how we define a norm.

So one more time: I suspect, that for a vector space with defined inner product, the norm of a vector, that lies in this vector space is also defined as induced by the inner product and is unambiguous. Is it true?

Answer 1

It will be possible to define multiple norms on any vector space (although they may be equivalent). For example, $||x|| = c\sqrt{\langle x , x \rangle }$ will also be a norm for any $c$. Indeed for your example, there is an infinite class of non-equivalent norms given by $$||P(x)||_p = \left ( \int_0^1\left |P(x)\right|^pdx\right )^\frac1p$$ for any $p \in \mathbb R, p \ge 1$.

However, in the definition of an angle as you have expressed it, the norm there must be the standard norm $||x|| = \sqrt{\langle x , x \rangle }$, and if we were to define a new norm, it would not necessarily be consistent with the inner product.

In short, when we work with an inner product space we usually assume that we're working with the standard norm. But that doesn't mean that there aren't other norms.