Proof of The Norm of a Real Inner Product Space

Throughout, $\|x\|\ge0$ and $\|x\|^{2}=\langle x,x\rangle$ for $x\in V$, since $\|x\|$ is the nonnegative real number whose square is $\langle x,x\rangle$ (Real Inner Product Spaces (vii), The k-th Root and the Square Root of a Nonnegative Real Number (i)); and $\sqrt{a^{2}}=a$ for real $a\ge0$ (Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (i)).

Bilinearity and the zero vector. By symmetry and linearity in the first argument (Real Inner Product Spaces (i), Real Inner Product Spaces (ii), Real Inner Product Spaces (iii)),

$$\langle u,v+w\rangle=\langle v+w,u\rangle=\langle v,u\rangle+\langle w,u\rangle=\langle u,v\rangle+\langle u,w\rangle,\qquad \langle u,\lambda v\rangle=\langle\lambda v,u\rangle=\lambda\,\langle v,u\rangle=\lambda\,\langle u,v\rangle .$$

Since $0_V=0\,0_V$ (Elementary Identities in a Vector Space (ii)), homogeneity gives $\langle 0_V,v\rangle=0\,\langle 0_V,v\rangle=0$, and $\langle v,0_V\rangle=0$ by symmetry. This proves (i).

Expansion of squares. Expanding in the first argument and then in the second ((i)), and using symmetry,

$$\|u+v\|^{2}=\langle u,u+v\rangle+\langle v,u+v\rangle=\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle=\|u\|^{2}+2\,\langle u,v\rangle+\|v\|^{2}.$$

For the difference, $u-v=u+(-1)v$ (The Zero Vector, Negatives and Differences in a Vector Space (iii), Elementary Identities in a Vector Space (iv)). Applying the first formula to $u$ and $(-1)v$, with $\langle u,(-1)v\rangle=-\langle u,v\rangle$ and $\|(-1)v\|^{2}=(-1)^{2}\langle v,v\rangle=\|v\|^{2}$ by linearity in both arguments ((i)), gives the second formula. This proves (ii).

The norm axioms. We check the conditions of Norms and Real Normed Spaces (v). Nonnegativity (Norms and Real Normed Spaces (i)) was noted at the start. Definiteness (Norms and Real Normed Spaces (ii)): $\|0_V\|=\sqrt{\langle 0_V,0_V\rangle}=\sqrt{0}=0$ by (i); conversely, if $\|v\|=0$, then $\langle v,v\rangle=\|v\|^{2}=0$, so $v=0_V$ by Real Inner Product Spaces (iv). Homogeneity (Norms and Real Normed Spaces (iii)): by linearity in both arguments ((i)) and the rules $\sqrt{xy}=\sqrt{x}\sqrt{y}$ for $x,y\ge0$ and $\sqrt{\lambda^{2}}=|\lambda|$ (Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (iii), Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (ii)),

$$\|\lambda v\|=\sqrt{\lambda^{2}\langle v,v\rangle}=\sqrt{\lambda^{2}}\,\sqrt{\langle v,v\rangle}=|\lambda|\,\|v\| .$$

Triangle inequality (Norms and Real Normed Spaces (iv)): $\langle u,v\rangle\le|\langle u,v\rangle|\le\|u\|\,\|v\|$ (Absolute Value in an Ordered Field (i), The Cauchy–Schwarz Inequality in a Real Inner Product Space (i)), so by (ii)

$$\|u+v\|^{2}\le\|u\|^{2}+2\,\|u\|\,\|v\|+\|v\|^{2}=\bigl(\|u\|+\|v\|\bigr)^{2},$$

and taking square roots, which preserves $\le$ between nonnegative numbers (Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (iv)), gives $\|u+v\|\le\|u\|+\|v\|$. As $V$ is a real vector space (Real Inner Product Spaces (vi)), $V$ with $\|\cdot\|$ is a real normed space (Norms and Real Normed Spaces (vi)). This proves (iii).

Parallelogram law. Adding the two formulas of (ii), the terms $\pm2\langle u,v\rangle$ cancel:

$$\|u+v\|^{2}+\|u-v\|^{2}=2\,\|u\|^{2}+2\,\|v\|^{2}.$$

Continuity. Since $d_V(x,y)=\|x-y\|$ (The Norm Metric of a Real Normed Space (i)), the convergence $u_m\to u$ in $(V,d_V)$ says that for every $\varepsilon>0$ there is $N\in\mathbb{N}$ with $\|u_m-u\|<\varepsilon$ for $m\ge N$ (Convergent Sequence in a Metric Space (i)); as $\|u_m-u\|\ge0$, this means $\|u_m-u\|\to0$ in $\mathbb{R}$ (Convergent Sequences of Real Numbers (i)). Likewise $\|v_m-v\|\to0$, and $\|v_m\|\to\|v\|$ (The Norm Metric of a Real Normed Space (vii)). By linearity in both arguments ((i)), with $x-y=x+(-1)y$ as above,

$$\langle u_m,v_m\rangle-\langle u,v_m\rangle=\langle u_m-u,v_m\rangle,\qquad \langle u,v_m\rangle-\langle u,v\rangle=\langle u,v_m-v\rangle .$$

The triangle inequality for the absolute value (Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (xi)) and the Cauchy–Schwarz inequality (The Cauchy–Schwarz Inequality in a Real Inner Product Space (i)) give, for every $m\in\mathbb{N}$,

$$\bigl|\langle u_m,v_m\rangle-\langle u,v\rangle\bigr|\le\bigl|\langle u_m-u,v_m\rangle\bigr|+\bigl|\langle u,v_m-v\rangle\bigr|\le\|u_m-u\|\,\|v_m\|+\|u\|\,\|v_m-v\|=:b_m .$$

By the arithmetic of limits, the constant sequence $(\|u\|)$ converging to $\|u\|$ (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i)), $b_m\to0\cdot\|v\|+\|u\|\cdot0=0$, so $(b_m)$ is a null sequence, and $\langle u_m,v_m\rangle\to\langle u,v\rangle$ by domination (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xii)). This proves (v).