Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation

Let $n\in \mathbb{N}$ and let $x,y\in {\mathbb{R}}^n$, with the Euclidean inner product and the Euclidean norm as in The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. Then:

1. Cauchy-Schwarz. $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2,$ with equality if and only if there is a pair $(\lambda ,\mu )\mathrm{\ne }(0,0)$ of reals with $\lambda x_k=\mu y_k$ for every $k<n$.
2. $\parallel \cdot {\parallel }_2$ is a norm on ${\mathbb{R}}^n$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), for every $n\in \mathbb{N}$; the metric it induces is $d_2$ of ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it whenever $n\ge 1$.
3. Parallelogram law. $\parallel x+y{\parallel }_2^2+\parallel x-y{\parallel }_2^2=2\parallel x{\parallel }_2^2+2\parallel y{\parallel }_2^2\mathrm{.}$
4. Polarisation. $⟨x,y⟩=\frac{1}{4}(\parallel x+y{\parallel }_2^2-\parallel x-y{\parallel }_2^2),$ so the inner product is recovered from the norm it induces.

Clause 1 is a citation, not a new proof. The inequality and its equality case are the published The Cauchy-Schwarz inequality for finite sums, stated there for two lists of reals; all that happens below is that it is read in the vector notation of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. Re-proving it here would put two proofs of one statement in the library.

• Which route to the triangle inequality was taken. The proof above obtains (N3) by expanding $\parallel x+y{\parallel }_2^2$ and applying Cauchy-Schwarz. The alternative is to quote Minkowski's inequality for finite sums (rational exponent) at the rational exponent $p=2$, which states the same inequality directly; that route is equally legitimate and is the one Each $\parallel \cdot {\parallel }_p$ is a norm on ${\mathbb{R}}^n$, and the induced metrics are exactly $d_1$, $d_2$ and $d_{\mathrm{\infty }}$ of the published metric-spaces page uses for a general exponent. Only one of the two is used here, so no statement is proved twice.

• Clause 1 holds at $n=0$, where it reads $0\le 0$, and the equality case is then satisfied by every pair $(\lambda ,\mu )\mathrm{\ne }(0,0)$, the condition quantifying over no indices. Clause 2 also holds at $n=0$, the zero space carrying exactly one norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). What is not available at $n=0$ is the metric $d_2$ of ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it, which is why the last sentence of clause 2 carries $n\ge 1$.

• Clauses 3 and 4 are what the companion page uses. The parallelogram law is an identity satisfied by every norm of the form $\sqrt{⟨\cdot ,\cdot ⟩}$, so a norm violating it is not of that form; that is how the companion page rules out $\parallel \cdot {\parallel }_1$ on ${\mathbb{R}}^2$. Polarisation says the inner product carries no information the norm does not.