Editing Inner product space (section)

==Notes==

{{reflist|group="Note"|refs=<!--
<ref group="Note" name=ConjugateNotation>A line over an expression or symbol, such as <math>\overline{s}</math> or <math>\overline{\langle y, x \rangle},</math> denotes [[Complex conjugate|complex conjugation]]. A scalar <math>s</math> is real if and only if <math>s = \overline{s}.</math></ref>

<ref group="Note" name=DefAsPosDefSesquilinear>This is because {{EquationNote|Additivity in the 1st argument|condition (1)}} (that is, linearity in the first argument) and {{EquationNote|Positive definite|positive definiteness}} implies that <math>\langle x, x \rangle</math> is always a real number. And as mentioned before, a sesquilinear form is Hermitian if and only if <math>\langle x, x \rangle</math> is real for all <math>x.</math></ref>

<ref group="Note" name=DefByPolarization>Let <math>R(x, y) := \frac{1}{4} \left(\|x + y\|^2 - \|x - y\|^2\right).</math> If <math>\mathbb{F} = \R</math> then let <math>\langle x,\, y \rangle_P := R(x, y)</math> while if <math>\mathbb{F} = \C</math> then let <math>\langle x,\, y \rangle_P := R(x, y) + i R(x, i y).</math> See the [[polarization identity]] article for more details.</ref>

<ref group="Note: name=SuggestsConjHom>If <math>\langle x,\, c y \rangle</math> can be written as <math>\langle x,\, c y \rangle = f(c, y) \langle x,\, y \rangle</math> for some function <math>f</math> (in particular, this assumes that the scalar in front of <math>\langle x,\, y \rangle</math> that results from trying to "pull <math>c</math> out of <math>\langle x,\, c y \rangle</math>" does not depend on <math>x</math>) then <math>\langle y,\, c y \rangle = \overline{c} \langle y,\, y \rangle</math> implies that <math>f(c, y) = \overline{c}</math> (when <math>y \neq 0</math>) and consequently, <math>\langle x,\, c y \rangle = \overline{c} \langle x,\, y \rangle</math> will hold for all <math>x, y, \text{ and } c.</math></ref>
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'''Proofs'''

{{reflist|group=proof|refs=
<ref group=proof name=ZeroVecProduces0AndRationalHomogeneousProof>{{EquationNote|Homogeneity in the 1st argument}} implies <math> \langle q x, y \rangle = 0 \langle x, y \rangle</math> for all rational <math>q</math> so that <math>\langle \mathbf{0}, y \rangle = \langle 0 y, y \rangle = 0 \langle y, y \rangle = 0.</math> <math>\blacksquare</math> Assume {{EquationNote|Additivity in the 1st argument|additivity in the 1st argument}}. Then <math>\langle \mathbf{0}, y \rangle = \langle \mathbf{0} + \mathbf{0}, y \rangle = \langle \mathbf{0}, y \rangle + \langle \mathbf{0}, y \rangle</math> so adding <math>- \langle \mathbf{0}, y \rangle</math> to both sides proves <math>\langle \mathbf{0}, y \rangle = 0.</math> Consequently, <math>0 = \langle \mathbf{0}, y \rangle = \langle x + (-x), y \rangle = \langle x, y \rangle + \langle -x, y \rangle,</math> which implies <math>\langle - x, y \rangle = - \langle x, y \rangle.</math> Induction shows that <math>\langle m x, y \rangle = m \langle x, y \rangle</math> for all integers <math>m.</math> If <math>n > 0</math> is an integer then <math>\langle x, y \rangle = \langle n \left(\tfrac{1}{n} x\right), y \rangle = n \langle \tfrac{1}{n} x, y \rangle</math> so that <math>\langle \tfrac{1}{n} x, y \rangle = \tfrac{1}{n} \langle x, y \rangle.</math> It follows that <math>\langle q x, y \rangle = q \langle x, y \rangle</math> for all rational <math>q \in \Q.</math> <math>\blacksquare</math> An analogous proof show that {{EquationNote|Additivity in the 2nd argument|additivity in the 2nd argument}} and {{EquationNote|Conjugate homogeneity in the 2nd argument|conjugate homogeneity in the 2nd argument}} each individually imply that <math>\langle x, q y \rangle = q \langle x, y \rangle</math> for all rational <math>q \in \Q.</math></ref>

<ref group=proof name=SesqHermEquivProof>Assume that it is a sesquilinear form that satisfies <math>\langle x, x \rangle \in \R</math> for all <math>x.</math> To conclude that <math>\langle x, y \rangle = \overline{\langle y, x \rangle},</math> it is necessary and sufficient to show that the real parts of <math>\langle y, x \rangle</math> and <math>\langle x, y \rangle</math> are equal and that their imaginary parts are negatives of each other. For all <math>x, y,</math> because <math>\langle x + y, x + y \rangle - \langle x, x \rangle - \langle y, y \rangle = \langle y, x \rangle + \langle x, y \rangle</math> and the left hand side is real, <math>\langle y, x \rangle + \langle x, y \rangle</math> is also real, which implies that the <math>0 = \operatorname{im} \left[\langle y, x \rangle + \langle x, y \rangle\right] = \left(\operatorname{im} \langle y, x \rangle\right) + \operatorname{im} \langle x, y \rangle.</math> Similarly, <math>\langle i y, x \rangle + \langle x, i y \rangle \in \R.</math> But sesquilinearity implies <math>\langle i y, x \rangle + \langle x, i y \rangle = i (\langle y, x \rangle - \langle x, y \rangle),</math> which is only possible if the real parts of <math>\langle y, x \rangle</math> and <math>\langle x, y \rangle</math> are equal. <math>\blacksquare</math></ref>

<ref group=proof name=HermSymImpliesRealProof>A complex number <math>c</math> is a real number if and only if <math>c = \overline{c}.</math> Using <math>y = x</math> in {{EquationNote|Conjugate symmetry|condition (2)}} gives <math>\langle x, x \rangle = \overline{\langle x, x \rangle},</math> which implies that <math>\langle x, x \rangle</math> is a real number. <math>\blacksquare</math></ref>

<ref group=proof name=BilinearRangeIsCProof>Assume that <math>\langle \,\cdot, \cdot\, \rangle</math> is a [[bilinear map]] and that <math>x \in V</math> satisfies <math>\langle x, x \rangle \neq 0.</math> Let <math>N : \mathbb{F} \to \mathbb{F}</math> be defined by <math>N(c) := \langle c x, c x \rangle</math> where bilinearity implies that <math>N(c) = \langle c x, c x \rangle = c^2 \langle x, x \rangle = c^2 N(1)</math> holds for all scalars <math>c.</math> Since <math>N(1) = \langle x, x \rangle \neq 0,</math> the scalar <math>1/N(1)</math> is well-defined and so <math>N(c) = 0</math> if and only if <math>c = 0.</math> If <math>c \in \Complex</math> is a scalar such that <math>c^2 \not\in \R</math> (which implies <math>c \neq 0</math> and <math>\frac{1}{c^2} \not\in \R</math>) then <math>N(1) \in \R</math> implies <math>N(c) = c^2 N(1) \not\in \R</math> and similarly, <math>N(c) \in \R</math> implies <math>N(1) = \frac{1}{c^2} N(c) \not\in \R;</math> this shows that for such a <math>c,</math> at most one of <math>N(1) \text{ and } N(c)</math> can be real. <math>\blacksquare</math> 
If <math>\mathbb{F} = \Complex</math> and <math>s \in \mathbb{F}</math> then pick <math>c \in \Complex</math> such that <math>c^2 = \frac{s}{N(1)},</math> which implies that <math>N(c) = c^2 N(1) = \frac{s}{N(1)} N(1) = s;</math> thus <math>N(\Complex) = \Complex</math> so <math>N : \Complex \to \Complex</math> is surjective. If <math>\mathbb{F} = \R</math> and <math>R(1) > 0</math> (resp. <math>R(1) < 0</math>) then for any <math>s \geq 0</math> (resp. any <math>s \leq 0</math>), <math>N\left(\sqrt{s/N(1)}\right) = s,</math> which shows that <math>N(\R) = [0, \infty)</math> (resp. <math>N(\R) = (-\infty, 0]</math>). <math>\blacksquare</math></ref>

<ref group=proof name=parallelogramLawSatisfiedProof>Note that <math>\|x+y\|^2 = \langle x+y, x+y\rangle = \langle x, x\rangle + \langle x, y\rangle + \langle y, x\rangle + \langle y, y\rangle</math> and <math>\|x-y\|^2 = \langle x-y, x-y\rangle = \langle x, x\rangle - \langle x, y\rangle - \langle y, x\rangle + \langle y, y\rangle,</math> which implies that <math>\|x+y\|^2 + \|x-y\|^2 = 2\langle x, x\rangle + 2\langle y, y\rangle = 2\|x\|^2 + 2\|y\|^2.</math> This proves that <math>\|\,\cdot\,\|</math> satisfies the [[parallelogram law]]. It also follows that <math>\|x+y\|^2 = \|x - y\|^2 + 2[\langle x, y \rangle + \langle y, x \rangle],</math> which proves that <math>\langle x, y \rangle + \langle y, x \rangle</math> is a real number and thus that its [[imaginary part]] is <math>0.</math> This implies that <math>\operatorname{im} \langle x, y \rangle = - \operatorname{im} \langle y, x \rangle.</math> If <math>\mathbb{F} = \Complex</math> then also <math>\langle x, iy \rangle + \langle iy, x \rangle = -[\langle ix, y \rangle + \langle y, ix \rangle].</math> <math>\blacksquare</math></ref>

<ref group=proof name=InnerProductOfxANDixProof>Combining <math>\|i x\| = |i| \|x\| = \|x\|</math> and <math>2\|x\|^2 = |1+i|^2 \, \|x\|^2 = \|(1+i)x\|^2 = \langle x + i x, x + i x \rangle = \|x\|^2 + \langle x, ix \rangle + \langle i x, x \rangle + \|ix\|^2</math> proves that <math>0 = \langle x, ix \rangle + \langle i x, x \rangle.</math> <math>\blacksquare</math></ref>

<ref group=proof name=RealHomIfContinuousProof>Fix <math>x, y \in V.</math> The equality <math>\langle q x, y \rangle = q \langle x, y \rangle</math> will be discussed first. Define <math>L, R : \R \to \mathbb{F}</math> by <math>L(q) := \langle q x, y \rangle</math> and <math>R(q) := q \langle x, y \rangle.</math> Because <math>\langle q x, y \rangle = q \langle x, y \rangle</math> for all <math>q \in \Q,</math> <math>L</math> and <math>R</math> are equal on a [[Dense set|dense subset]] of <math>\R.</math> Since <math>\langle x, y \rangle</math> is constant, the map <math>R : \R \to \mathbb{F}</math> is continuous (where the [[Hausdorff space]] <math>\mathbb{F},</math> which is either <math>\R</math> or <math>\Complex,</math> has its usual [[Euclidean topology]]). Consequently, if <math>L : \R \to \mathbb{F}</math> is also continuous then <math>L</math> and <math>R</math> will necessarily be equal on all of <math>\R;</math> that is, <math>\langle q x, y \rangle = q \langle x, y \rangle</math> will hold for all {{em|real}} <math>q \in \R.</math> If <math>f : \R \to V \text{ and } g : V \to \mathbb{F}</math> are defined by <math>f(q) := q x</math> and <math>g(v) := \langle v, y \rangle</math> then <math>L = g \circ f.</math> So for <math>L</math> to be continuous, it suffices for there to exist some topology <math>\tau</math> on <math>V</math> that makes both <math>f</math> and <math>g</math> continuous (or even just [[sequentially continuous]]). The map <math>f : \R \to (V, \tau)</math> will automatically be continuous if <math>\tau</math> is a [[topological vector space]] topology, such as a topology induced by a norm. The map <math>g : (V, \tau) \to \mathbb{F}</math> will be continuous if <math>\langle \,\cdot, \cdot\, \rangle : V \times V \to \mathbb{F}</math> is [[separately continuous]] (which will be true if <math>\langle \,\cdot, \cdot\, \rangle</math> is continuous). The discussion of the equality <math>\langle x, q y \rangle = q \langle x, y \rangle</math> is nearly identical, with the main difference being that <math>L, f, g</math> must be redefined as <math>L(q) := \langle x, q y \rangle,</math> <math>f(q) := q y,</math> and <math>g(v) := \langle x, v \rangle.</math> <math>\blacksquare</math></ref>

<ref group=proof name=LinAndHermSymImplyAntilinearProof>Let <math>x, y, z</math> be vectors and let <math>s</math> be a scalar. Then <math>\langle x, s y \rangle = \overline{\langle s y, x \rangle} = \overline{s} \overline{\langle y, x \rangle} = \overline{s} \langle x, y \rangle</math> and <math>\langle x, y + z \rangle = \overline{\langle y + z, x \rangle} = \overline{\langle y, x \rangle} + \overline{\langle z, x \rangle} = \langle x, y \rangle + \langle x, z \rangle.</math> <math>\blacksquare</math></ref>
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