--- title: Inner Product Spaces parent: Functional Analysis Basics nav_order: 6 has_children: true has_toc: false --- # {{ page.title }} {% definition Inner Product Space %} An *inner product* (or *scalar product*) on a real or complex vector space $X$ is a mapping $$ \innerp{\cdot}{\cdot} : X \times X \to \KK $$ that is - linear in its second argument $$ \innerp{x}{y+z} = \innerp{x}{y} + \innerp{x}{z} \qquad \innerp{x}{\alpha y} = \alpha \innerp{x}{y} $$ - conjugate symmetric $$ \overline{\innerp{x}{y}} = \innerp{x}{y} $$ - nondegenerate An *inner product space* (or *pre-Hilbert space*) is a pair $(X,\innerp{\cdot}{\cdot})$ consisting of a real or complex vector space $X$ and an inner product $\innerp{\cdot}{\cdot}$ on $X$. {% enddefinition %} {% proposition Norm Induced by an Inner Product %} If $\innerp{\cdot}{\cdot}$ is an inner product on a real or complex vector space $X$, then $$ \norm{x} = \sqrt{\innerp{x}{x}} \qquad \forall x \in X $$ defines a norm on $X$. {% endproposition %} In this sense, every inner product space is also a normed space. As a consequence it is also a metric space and a topological space. The next theorem shows how the inner product can be recovered from the norm. {% theorem * Polarization Identity %} For all vectors $x$ and $y$ of a real inner product space $$ 4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2. $$ For all vectors $x$ and $y$ of a complex inner product space $$ 4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2 + i \norm{x-iy}^2 - i \norm{x+iy}^2. $$ {% endtheorem %} Note that the complex polarization identity takes the slightly different form $$ 4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2 + i \norm{x\mathrel{\color{red}+}iy}^2 - i \norm{x\mathrel{\color{red}-}iy}^2, $$ if we follow the convention that the inner product is conjugate linear in its second argument. {% proof %} In the real case, the inner product is symmetric, and we have $$ \norm{x \pm y}^2 = \norm{x}^2 \pm 2 \innerp{x}{y} + \norm{y}^2 $$ for all vectors $x$ and $y$. Taking the difference yields the desired result. In the complex case, the inner product is conjugate symmetric, and we have $$ \norm{x \pm y}^2 = \norm{x}^2 \pm 2 \Re \innerp{x}{y} + \norm{y}^2 $$ for all vectors $x$ and $y$. This implies $$ \begin{aligned} \norm{x + \phantom{i}y}^2 - \norm{x - \phantom{i}y}^2 &= 4 \Re \innerp{x}{y}, \\ \norm{x - iy}^2 - \norm{x + iy}^2 &= 4 \Im \innerp{x}{y}. \end{aligned} $$ The second equation follows from the first by substituting $y$ with $-iy$ and using that $\Re \innerp{x}{-iy} = \Re (-i\innerp{x}{y}) = \Im \innerp{x}{y}$. To obtain the polarization Identity, multiply the second equation with $i$ and then add it to the first. {% endproof %} {% theorem * General Polarization Identity %} Let $X$ be a complex inner product space. Let $\zeta$ be a $n$-th root of unity with $\zeta \ne 1$ and $\zeta^2 \ne 1$. Then $$ \innerp{x}{y} = \frac{1}{n} \sum_{k=0}^{n-1} \zeta^k \norm{x + \zeta^k y}^2 \qquad \forall x,y \in X. $$ {% endtheorem %} As a special case, for $\zeta = i$ and $n=4$, we obtain $$ \innerp{x}{y} = \frac{1}{4} \sum_{k=0}^{3} i^k \norm{x + i^k y}^2. $$ {% proof %} TODO {% endproof %} For an arbitrary normed space, the polarization identity does not, in general, define an inner product. The following theorem, gives a condition for when it does. {% theorem * Parallelogram Law %} Let $X$ be a real or complex normed space. A norm $\norm{\cdot}$ on $X$ is induced by an inner product $\innerp{\cdot}{\cdot}$ on $X$, if and only if $\norm{\cdot}$ satisfies the *parallelogram law* $$ \norm{x+y}^2 + \norm{x-y}^2 = 2 \norm{x}^2 + 2 \norm{y}^2 \qquad \forall x,y \in X. $$ In this case, the inner product is uniquely determined by $\norm{\cdot}$ and given by the polarization identity. {% endtheorem %} {% theorem * Stewart’s Theorem %} Let $x$, $y$, $z$ be vectors of an inner product space. If $x$, $y$ and $z$ are colinear and $y$ lies inbetween $x$ and $y$, then we have $$ \norm{p-x}^2 \norm{y-z} + \norm{p-z}^2 \norm{x-y} = \big\lparen \norm{p-y}^2 + \norm{x-y} \norm{y-z} \big\rparen \norm{x-z} $$ {% endtheorem %} --- {% theorem * Cauchy–Schwarz Inequality %} For all vectors $x$ and $y$ of an inner product space (with inner product $\innerp{\cdot}{\cdot}$ and induced norm $\norm{\cdot}$) $$ \abs{\innerp{x}{y}} \le \norm{x} \norm{y}, $$ and equality holds precisely when $x$ and $y$ are linearly dependent. {% endtheorem %} Expressed only in terms of the inner product, the Cauchy–Schwarz Inequality reads $$ \abs{\innerp{x}{y}}^2 \le \innerp{x}{x} \innerp{y}{y}. $$ {% proof %} TODO {% endproof %} {% corollary Continuity of the Inner Product %} The inner product is jointly norm continuous. {% endcorollary %}