A calculus of the absurd

Every inner product “induces” a norm, which is to say that if we have an inner product, we can define the norm

\begin{equation} \norm {\mathbf {x}} = \sqrt {\langle \mathbf {x}, \mathbf {x} \rangle } \end{equation}

To show that this actually is a norm it is necessary and sufficient to show that it satisfies the four norm axioms.

• Homogeneity. Consider \(\norm {\alpha \mathbf {x}}\), for which
\(\seteqnumber{0}{22.}{155}\)
\begin{align} \norm {\alpha \mathbf {x}} &= \sqrt {\langle \alpha \mathbf {x}, \alpha \mathbf {x} \rangle } \\ &= \sqrt {\alpha \langle \mathbf {x}, \alpha \mathbf {x} \rangle } \\ &= \sqrt {\alpha \overline {\langle \alpha \mathbf {x}, \mathbf {x} \rangle }} \\ &= \sqrt {\alpha \overline {\alpha } \overline {\langle \mathbf {x}, \mathbf {x} \rangle }} \\ &= \sqrt {\abs {\alpha } \langle \mathbf {x}, \mathbf {x} \rangle } \end{align}
• TODO: other proofs

We will pretend these do not exist (but be aware that they do most definitely exist)!!!