Canonical norm

Inner Products $⟨\cdot ,\cdot ⟩$ define a canonical $p:X\to \mathbb{R}$ via $p(x):=\sqrt{⟨x,x⟩}$ . We can check that this satisfies all of the Norm axioms:

1. $p(x+y{)}^2=⟨x+y,x+y⟩=⟨x,x⟩+⟨y,y⟩+⟨x,y⟩+⟨y,x⟩=p(x{)}^2+p(y{)}^2+2|⟨x,y⟩|\le p(x{)}^2+p(y{)}^2+2p(x)p(y)=(p(x)+p(y){)}^2$, where the inequality is the result of using the Cauchy-Schwarz inequality . Taking square roots, this gives us the triangle inequality.
2. $p(sx)=\sqrt{⟨sx,sx⟩}=\sqrt{s\overline{s}⟨x,x⟩}=\Vert s\Vert s\sqrt{x,x}=\Vert s\Vert p(x)$, showing us that we have absolute homogeneity
3. $p(x)=0\Rightarrow \sqrt{⟨x,x⟩}=0\Rightarrow ⟨x,x⟩=0\Rightarrow x=0$, giving us point separation

• Clarify the proofs above using block maths