Editing Riesz representation theorem (section)

== Notes ==

{{reflist|group=note|refs=
<ref group=note name="AntilinearIsometryDef">This means that for all vectors <math>y \in H:</math> (1) <math>\Phi : H \to H^*</math> is [[injective]]. (2) The [[Norm (mathematics)|norms]] of <math>y</math> and <math>\Phi(y)</math> are the same: <math>\|\Phi(y)\| = \|y\|.</math> (3) <math>\Phi</math> is an [[additive map]], meaning that <math>\Phi(x + y) = \Phi(x) + \Phi(y)</math> for all <math>x, y \in H.</math> (4) <math>\Phi</math> is [[conjugate homogeneous]]: <math>\Phi(s y) = \overline{s} \Phi(y)</math> for all scalars <math>s.</math> (5) <math>\Phi</math> is [[real homogeneous]]: <math>\Phi(r y) = r \Phi(y)</math> for all real numbers <math>r \in \R.</math></ref>

<ref group=note name="VectorSpaceStructureOnAffineHyperplanesInducedByDualSpace">This footnote explains how to define - using only <math>H</math>'s operations - addition and scalar multiplication of affine hyperplanes so that these operations correspond to addition and scalar multiplication of linear functionals. Let <math>H</math> be any vector space and let <math>H^{\#}</math> denote its [[algebraic dual space]]. Let <math>\mathcal{A} := \left\{ \varphi^{-1}(1) : \varphi \in H^{\#} \right\}</math> and let <math>\,\hat{\cdot}\,</math> and <math>\,\hat{+}\,</math> denote the (unique) vector space operations on <math>\mathcal{A}</math> that make the bijection <math>I : H^{\#} \to \mathcal{A}</math> defined by <math>\varphi \mapsto \varphi^{-1}(1)</math> into a [[vector space isomorphism]]. Note that <math>\varphi^{-1}(1) = \varnothing</math> if and only if <math>\varphi = 0,</math> so <math>\varnothing</math> is the additive identity of <math>\left(\mathcal{A}, \hat{+}, \hat{\cdot}\right)</math> (because this is true of <math>I^{-1}(\varnothing) = 0</math> in <math>H^{\#}</math> and <math>I</math> is a vector space isomorphism). For every <math>A \in \mathcal{A},</math> let <math>\ker A = H</math> if <math>A = \varnothing</math> and let <math>\ker A = A - A</math> otherwise; if <math>A = I(\varphi) = \varphi^{-1}(1)</math> then <math>\ker A = \ker \varphi</math> so this definition is consistent with the usual definition of the kernel of a linear functional. Say that <math>A, B \in \mathcal{A}</math> are {{em|parallel}} if <math>\ker A = \ker B,</math> where if <math>A</math> and <math>B</math> are not empty then this happens if and only if the linear functionals <math>I^{-1}(A)</math> and <math>I^{-1}(B)</math> are non-zero scalar multiples of each other. The vector space operations on the vector space of affine hyperplanes <math>\mathcal{A}</math> are now described in a way that involves {{em|only}} the vector space operations on <math>H</math>; this results in an interpretation of the vector space operations on the algebraic dual space <math>H^{\#}</math> that is entirely in terms of affine hyperplanes. Fix hyperplanes <math>A, B \in \mathcal{A}.</math> If <math>s</math> is a scalar then <math>s \hat{\cdot} A = \left\{ h \in H : s h \in A \right\}.</math> Describing the operation <math>A \hat{+} B</math> in terms of only the sets <math>A = \varphi^{-1}(1)</math> and <math>B = \psi^{-1}(1)</math> is more complicated because by definition, <math>A \hat{+} B = I(\varphi) \hat{+} I(\psi) := I(\varphi + \psi) = (\varphi + \psi)^{-1}(1).</math> If <math>A = \varnothing</math> (respectively, if <math>B = \varnothing</math>) then <math>A \hat{+} B</math> is equal to <math>B</math> (resp. is equal to <math>A</math>) so assume <math>A \neq \varnothing</math> and <math>B \neq \varnothing.</math> The hyperplanes <math>A</math> and <math>B</math> are parallel if and only if there exists some scalar <math>r</math> (necessarily non-0) such that <math>A = r B,</math> in which case <math>A \hat{+} B = \left\{ h \in H : (1 + r) h \in B \right\};</math> this can optionally be subdivided into two cases: if <math>r = -1</math> (which happens if and only if the linear functionals <math>I^{-1}(A)</math> and <math>I^{-1}(B)</math> are negatives of each) then <math>A \hat{+} B = \varnothing</math> while if <math>r \neq -1</math> then <math>A \hat{+} B = \frac{1}{1+r} B = \frac{r}{1+r} A.</math> Finally, assume now that <math>\ker A \neq \ker B.</math> Then <math>A \hat{+} B</math> is the unique affine hyperplane containing both <math>A \cap \ker B</math> and <math>B \cap \ker A</math> as subsets; explicitly, <math>\ker \left(A \hat{+} B\right) = \operatorname{span}\left(A \cap \ker B - B \cap \ker A\right)</math> and <math>A \hat{+} B = A \cap \ker B + \ker \left(A \hat{+} B\right) = B \cap \ker A + \ker \left(A \hat{+} B\right).</math> To see why this formula for <math>A \hat{+} B</math> should hold, consider <math>H := \R^3,</math> <math>A := \varphi^{-1}(1),</math> and <math>B := \psi^{-1}(1),</math> where <math>\varphi(x, y, z) := x</math> and <math>\psi(x, y, z) := x + y</math> (or alternatively, <math>\psi(x, y, z) := y</math>). Then by definition, <math>A \hat{+} B := (\varphi + \psi)^{-1}(1)</math> and <math>\ker \left(A \hat{+} B\right) := (\varphi + \psi)^{-1}(0).</math> Now <math>A \cap \ker B ~=~ \varphi^{-1}(1) \cap \psi^{-1}(0) ~\subseteq~ (\varphi + \psi)^{-1}(1)</math> is an affine subspace of [[codimension]] <math>2</math> in <math>H</math> (it is equal to a translation of the <math>z</math>-axis <math>\{(0, 0)\} \times \R</math>). The same is true of <math>B \cap \ker A.</math> Plotting an <math>x</math>-<math>y</math>-plane cross section (that is, setting <math>z = </math> constant) of the sets <math>\ker A, \ker B, A</math> and <math>B</math> (each of which will be plotted as a line), the set <math>(\varphi + \psi)^{-1}(1)</math> will then be plotted as the (unique) line passing through the <math>A \cap \ker B</math> and <math>B \cap \ker A</math> (which will be plotted as two distinct points) while <math>(\varphi + \psi)^{-1}(0) = \ker \left(A \hat{+} B\right)</math> will be plotted the line through the origin that is parallel to <math>A \hat{+} B = (\varphi + \psi)^{-1}(1).</math> The above formulas for <math>\ker \left(A \hat{+} B\right) := (\varphi + \psi)^{-1}(0)</math> and <math>A \hat{+} B := (\varphi + \psi)^{-1}(1)</math> follow naturally from the plot and they also hold in general.</ref>

<ref group=note name="ImportanceOfLocationOfRieszRep">If <math>\mathbb{F} = \R</math> then the inner product will be symmetric so it does not matter which coordinate of the inner product the element <math>y</math> is placed into because the same map will result. 
But if <math>\mathbb{F} = \Complex</math> then except for the constant <math>0</math> map, [[Antilinear map|antilinear functionals]] on <math>H</math> are completely distinct from [[linear functional]]s on <math>H,</math> which makes the coordinate that <math>y</math> is placed into is {{em|very}} important. 
For a non-zero <math>y \in H</math> to induce a {{em|linear}} functional (rather than an {{em|anti}}linear functional), <math>y</math> {{em|must}} be placed into the {{em|anti}}linear coordinate of the inner product. If it is incorrectly placed into the linear coordinate instead of the antilinear coordinate then the resulting map will be the antilinear map <math>h \mapsto \langle y , h \rangle = \langle h \mid y \rangle,</math> which is {{em|not}} a linear functional on <math>H</math> and so it will {{em|not}} be an element of the continuous dual space <math>H^*.</math></ref>

<ref group=note name="ExplicitDefOfInnerProductOfTranspose">The notation <math>\left\langle z \mid A(\cdot) \right\rangle_Z</math> denotes the continuous linear functional defined by <math>g \mapsto \left\langle z \mid A g \right\rangle_Z.</math></ref>
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'''Proofs'''

{{reflist|group=proof|refs=
<ref group=proof name="FormulaOrthoProjectionKernel">This is because <math>x_K = x - \frac{\left\langle x, f_\varphi \right\rangle}{\left\|f_\varphi\right\|^2} f_{\varphi}.</math> Now use <math>\left\| f_{\varphi} \right\|^2 = \|\varphi\|^2</math> and <math>\left\langle x, f_{\varphi} \right\rangle = \varphi(x)</math> and solve for <math>f_{\varphi}.</math> <math>\blacksquare</math></ref> 

<ref group=proof name="NormalCharFunctionals"><math>\left\langle A^* A z \mid h \right\rangle = \left\langle \,A z\mid A h\, \right\rangle_H = \left\langle \,\Phi A h\mid\Phi A z\, \right\rangle_{H^*}</math> where <math>\Phi A h := \left\langle A h\mid\cdot\, \right\rangle</math> and <math>\Phi A z := \left\langle A z \mid\cdot\, \right\rangle.</math> 
By definition of the adjoint, <math>\left\langle A^* h\mid A^* z\, \right\rangle = \left\langle h \mid A A^* z\, \right\rangle</math> so taking the complex conjugate of both sides proves that <math>\left\langle A A^* z\mid h \right\rangle = \left\langle A^* z\mid A^* h \right\rangle.</math> 
From <math>A^* = \Phi^{-1} \circ {}^{t}A \circ \Phi,</math> it follows that <math>\left\langle A A^* z\, | \,h \right\rangle_H 
= \left\langle A^* z\mid A^* h \right\rangle_H 
= \left\langle \Phi^{-1} \circ {}^{t}A \circ \Phi z\mid\Phi^{-1} \circ {}^t A \circ \Phi h \right\rangle_H 
= \left\langle \,{}^{t}A \circ \Phi h\mid{}^t A \circ \Phi z \right\rangle_{H^*}</math> where <math>\left({}^{t}A \circ \Phi\right) h = \langle h\, | \,A (\cdot) \rangle</math> and <math>\left({}^{t}A \circ \Phi\right) z = \langle z\, | \,A (\cdot) \rangle.</math> <math>\blacksquare</math></ref>
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