Proof of Elementary Identities in a Real Inner Product Space

Throughout, (a), (b), (c), (d) refer to the conditions of Real Inner Product Spaces (i), and $|x|$ is the unique real number $r$ with $0\le r$ and $r^{2}=\langle x,x\rangle$ provided by Existence and Uniqueness of the Nonnegative Square Root as in Real Inner Product Spaces (ii). We use freely the field axioms of $\mathbb{R}$ and the vector space axioms of Vector Space over a Field.

Claim 1. By (a), (b) and (a) again, $\langle x,y+z\rangle=\langle y+z,x\rangle=\langle y,x\rangle+\langle z,x\rangle=\langle x,y\rangle+\langle x,z\rangle$; by (a), (c) and (a), $\langle x,\lambda y\rangle=\langle\lambda y,x\rangle=\lambda\langle y,x\rangle=\lambda\langle x,y\rangle$. By claim 5 of Elementary Identities in a Vector Space, $-x=(-1)x$, so by (c), $\langle -x,y\rangle=(-1)\langle x,y\rangle=-\langle x,y\rangle$, and by symmetry (a) also $\langle x,-y\rangle=-\langle x,y\rangle$. Finally $\langle x-y,z\rangle=\langle x+(-y),z\rangle=\langle x,z\rangle+\langle -y,z\rangle=\langle x,z\rangle-\langle y,z\rangle$ by (b) and what was just shown, and likewise $\langle x,y-z\rangle=\langle x,y\rangle-\langle x,z\rangle$ by the additivity in the second argument just proved.

Claim 2. By claim 3 of Elementary Identities in a Vector Space, $0_{E}=0\,x$, so by (c), $\langle 0_{E},x\rangle=\langle 0\,x,x\rangle=0\,\langle x,x\rangle=0$, and $\langle x,0_{E}\rangle=0$ by (a). In particular $\langle 0_{E},0_{E}\rangle=0=0^{2}$ with $0\le 0$, so $|0_{E}|=0$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root.

Claim 3. If $|x|=0$ then $\langle x,x\rangle=|x|^{2}=0$, so $x=0_{E}$ by (d). Conversely $|0_{E}|=0$ by claim 2. If $x\ne 0_{E}$, then $|x|\ne 0$ by what was just shown, while $0\le|x|$; hence $0<|x|$ by the definition of the strict order.

Claim 4. By (c) and claim 1, $\langle\lambda x,\lambda x\rangle=\lambda\langle x,\lambda x\rangle=\lambda^{2}\langle x,x\rangle$. By claim 1 of Properties of the Absolute Value in an Ordered Field, $|\lambda|$ equals $\lambda$ or $-\lambda$, and $(-\lambda)^{2}=\lambda^{2}$ in the field $\mathbb{R}$, so $|\lambda|^{2}=\lambda^{2}$. Hence $(|\lambda|\,|x|)^{2}=|\lambda|^{2}|x|^{2}=\lambda^{2}\langle x,x\rangle=\langle\lambda x,\lambda x\rangle=|\lambda x|^{2}$. Both $|\lambda x|$ and $|\lambda|\,|x|$ are nonnegative: the first by Real Inner Product Spaces (ii), the second because $0\le|\lambda|$ (claim 1 of Properties of the Absolute Value in an Ordered Field) and $0\le|x|$, so that $0=|\lambda|\cdot 0\le|\lambda|\,|x|$ by claim 5 of Elementary Arithmetic in an Ordered Field. Hence they are equal by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. For the last assertion, $-x=(-1)x$, and $|-1|=|1|$ by claim 2 of Properties of the Absolute Value in an Ordered Field while $|1|=1$ by Absolute Value in an Ordered Field since $0\le 1$ (claim 1 of Elementary Arithmetic in an Ordered Field); so $|-x|=|-1|\,|x|=|x|$.

Claim 5. By (b) and claim 1,

$$|x+y|^{2}=\langle x+y,x+y\rangle=\langle x,x+y\rangle+\langle y,x+y\rangle=\langle x,x\rangle+\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle=|x|^{2}+2\langle x,y\rangle+|y|^{2},$$

using (a) for $\langle y,x\rangle=\langle x,y\rangle$. Applying this with $-y$ in place of $y$, and using $\langle x,-y\rangle=-\langle x,y\rangle$ (claim 1) and $|-y|=|y|$ (claim 4), gives $|x-y|^{2}=|x|^{2}-2\langle x,y\rangle+|y|^{2}$.

Claim 6. Add the two identities of claim 5.

Claim 7. Subtract the second identity of claim 5 from the first: $|x+y|^{2}-|x-y|^{2}=4\langle x,y\rangle$.