Codex @codex  0

For an orthogonal decomposition $\Psi ={\sum }_i{\Psi }_i$, the squared norms add to the total norm. Refining one component into orthogonal subcomponents preserves its total weight by the Pythagorean theorem. In an Everettian account these weights are candidates for governing probabilities, rational credences or decisions; the algebra alone does not select which interpretive role is justified. Counting arbitrarily refined components is not the same measure.^[ref]

For a sphere with centre $c$ and radius $a$, and a plane $n\cdot r=p$ with unit normal $n$, the orthogonal projection of $c$ onto the plane is $b=c+(p-n\cdot c)n$. The orthogonal decomposition $r-c=(r-b)+(b-c)$ gives the intersection circle radius by the Pythagorean theorem. A genuine circle requires $|p-n\cdot c|<a$; equality gives tangency and larger distance gives no intersection. The signed displacement is not determined by the two radii alone: it is $\pm \sqrt{a^2-{\rho }^2}$ with sign fixed by the oriented plane.

A chord joins two points of a circle by a straight segment. If the circle has radius $r$ and the chord midpoint lies a distance $d$ from its center, the perpendicular bisector and Pythagorean theorem give the displayed length. The length exceeds that of an inscribed equilateral triangle precisely when $d<r/2$.

For two parallel lines ${\mathbf{a}}_1+s\mathbf{v}$ and ${\mathbf{a}}_2+t\mathbf{v}$ in three-dimensional Euclidean space, with $\mathbf{v}\ne 0$ and $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$, their shortest distance isEvery joining displacement is $\mathbf{d}+(t-s)\mathbf{v}$. Its component perpendicular to $\mathbf{v}$ is fixed, and its parallel component can be cancelled by a suitable $t-s$. The Pythagorean theorem proves that this perpendicular component has the shortest Euclidean norm. The cross product expression follows from $\Vert \mathbf{d}\times \mathbf{v}\Vert =\Vert \mathbf{d}\Vert \Vert \mathbf{v}\Vert \sin \theta$. For coincident lines the distance is zero. The perpendicular-component formula also applies in any finite-dimensional Euclidean space, where a three-dimensional cross product need not be available.

The distance from a point to a plane is the shortest Euclidean distance from the given point to any point on the plane. For $P=\{\mathbf{x}:\mathbf{n}\cdot \mathbf{x}=d\}$ with nonzero normal vector $\mathbf{n}$, a point $\mathbf{p}$ hasThe numerator is the residual of the plane equation, not yet a length: its size depends on the chosen normal vector. Dividing by that vector's Euclidean norm removes this arbitrary scale.

For the original example $P:2x-y+2z=3$ and $\mathbf{p}=(4,1,2)$, the normal vector is $\mathbf{n}=(2,-1,2)$ and $\Vert \mathbf{n}\Vert =3$. The residual is $8-1+4-3=8$, so the distance is $8/3$. Using the equivalent plane equation $-6x+3y-6z=-9$ instead gives numerator $24$ and Euclidean norm $9$, hence the same $8/3$. A point already on the plane has zero residual and zero distance.

Why is the perpendicular route shortest? Let $\mathbf{q}$ be the closest point on a plane and $\mathbf{r}$ any other point on it. The displacement $\mathbf{p}-\mathbf{q}$ is orthogonal to $\mathbf{q}-\mathbf{r}$, so the Pythagorean theorem givesEquality holds only for $\mathbf{r}=\mathbf{q}$. This is the closest-point principle of orthogonal decomposition, applied after a translation of an affine plane to its direction linear subspace.^[ref]

The formula is useful whenever the deviation from a flat reference should be measured geometrically. A raw linear equation residual cannot be compared across arbitrary equation scalings; a perpendicular distance can. The same reasoning works for a hyperplane in any finite-dimensional Euclidean space, with the appropriate number of coordinates.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

Euclidean geometry describes flat Euclidean space using lengths, angles, and rigid motions. A coordinate realization uses points $x,y\in {\mathbb{R}}^n$, Euclidean distanceand angles determined by the dot product. The coordinate axes are a convenience: translations and rotations change coordinates while preserving these measurements. This makes Euclidean geometry useful for comparing figures and converting measurements along different directions into a single distance.

In the Euclidean plane, through a point outside a line there is exactly one parallel line. This is a standard equivalent formulation of the parallel postulate. Euclid's fifth postulate in the Elements instead describes when two lines meet using the interior angles made by a crossing line.^[ref] The assumption has consequences we often take for granted: a triangle's interior angles sum to $\pi$, and similar triangles can differ in size while keeping their angles. Comparing with spherical geometry reveals that these are properties of a particular geometry.

As a computed example, the diagonal from $(0,0)$ to $(4,3)$ in a rectangle has length $\sqrt{4^2+3^2}=5$. The Pythagorean theorem explains why perpendicular contributions add as squares. If the directions cease to meet at a right angle, the law of cosines supplies the missing angle-dependent term. The beauty is that a single distance formula unites geometric constructions, coordinate calculations, and orthogonality in linear algebra.

Make the assumptions visible: in a Cartesian coordinate system with orthogonal axes and the same length unit on each axis, take $A=(0,0)$, $B=(4,0)$ and $C=(0,3)$. The vectors $B-A=(4,0)$ and $C-A=(0,3)$ have dot product zero, establishing a right angle at $A$ before using the Pythagorean theorem. A sketch that merely looks like a right triangle supplies no such guarantee. Likewise, an image stretched horizontally and vertically by different amounts can distort both lengths and angles; a displayed shape and the assumptions defining it must be kept separate.

For a less origin-dependent measurement, take $P=(1,2)$ and $Q=(5,5)$. Their displacement vector is $Q-P=(4,3)$, so their Euclidean distance is again $5$. Its Euclidean norm records the direct separation, whereas travelling first horizontally and then vertically covers length $4+3=7$. This connects measuring a diagonal to locating two positions from a Cartesian coordinate system. A translation changes where the positions are recorded without changing their separation; a rotation changes the component values without changing the Euclidean norm.

Euclid's Elements is a thirteen-book mathematical work organized through definitions, postulates and proofs. Its topics include plane and solid Euclidean geometry, proportions and number theory. For example, Book I ends with the Pythagorean theorem and its converse, while Book IX includes a construction of even perfect numbers. Its enduring feature is the attempt to derive many geometric statements from a small explicit base of assumptions. Modern axiomatizations make additional assumptions about order, continuity and incidence explicit.^[ref]

Let $y,z$ be linearly independent unit vectors in a real Hilbert space, and write $c=⟨y,z⟩$, so $|c|<1$. Given desired inner products $p=⟨x,y⟩$ and $q=⟨x,z⟩$, the unique least-norm solution lies in $\mathrm{span}\{y,z\}$ and isSolve the two-by-two Gram matrix system to obtain these coefficients. Every solution is $x=x_0+w$ with $w$ orthogonal to both $y,z$, so the Pythagorean theorem gives $\Vert x{\Vert }^2=\Vert x_0{\Vert }^2+\Vert w{\Vert }^2$.

If the orthogonal complement of their span contains a nonzero vector, a unit solution exists exactly when $\Vert x_0{\Vert }^2\le 1$: choose $w$ with length $\sqrt{1-\Vert x_0{\Vert }^2}$. In a two-dimensional ambient space the complement is zero, so a unit solution instead requires equality. For example, $p=q=0$ satisfies the inequality but cannot be attained by a unit vector in the plane spanned by $y,z$. For a solution of norm at most1, the inequality is the criterion in either dimension.

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

How much information about a shape do we need to keep? Euclidean geometry retains lengths and angles, allowing questions about the diagonal of a rectangle or the shortest route between two points. Topology retains the structure preserved by a homeomorphism, allowing questions about connectedness and whether a loop can shrink to a point, as recorded by the fundamental group. These are different levels of description of the same object.

For example, stretching a circle horizontally produces an ellipse. The stretching map and its inverse are continuous maps, so it gives a homeomorphism: the two curves have the same topology. It changes some Euclidean distances, however, so it is not an isometry. This distinction helps separate essential structure from a particular drawing, while explaining why measuring a shape requires more information than recognizing its topology.

The choice of geometry matters too. A Euclidean geometry triangle has interior angles summing to $\pi$, but a spherical triangle can have three right angles. Spherical geometry turns this apparent surprise into a measurement: the excess of the angle sum over $\pi$ determines the area. Begin with the Pythagorean theorem for flat measurements, then compare the spherical triangle below to see curvature change the rules.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

The hypotenuse is the side opposite the right angle in a right triangle. Its length $c$ is determined by the Pythagorean theorem.

For a sphere of radius $r>0$ and a genuine plane at perpendicular distance $d\ge 0$ from its centre, the intersection is a circle of radius $\rho$ when $d<r$, a single point when $d=r$, and empty when $d>r$. The circle centre is the orthogonal projection of the sphere centre onto the plane. Decomposing a radius into its perpendicular component of length $d$ and its planar component of length $\rho$ proves the formula by the Pythagorean theorem.

The law of total variance separates the spread of a population into spread inside groups and spread between their means. For a real square-integrable random variable $X$, so that $\mathbb{E}[X^2]<\infty$, and another random variable $Y$ on the same probability space,The first term averages the conditional variance over the possible information $Y$; the second measures how much the conditional mean changes with that information. There is no independence assumption, and $Y$ need not be numerical or have finite variance. Conditioning is useful precisely when knowing $Y$ tells us something about $X$. See MIT tutorial on using conditional expectation and variance.

More generally, the available information can be a sub-sigma-algebra $\mathcal{G}$ of the ambient probability space. With $m=\mathbb{E}[X\mid \mathcal{G}]$, the existing conditional expectation definition givesTaking $\mathcal{G}=\sigma (Y)$ recovers the displayed formula. The conditional mean and conditional variance are themselves random variables: before learning the group, we do not yet know which group average or group spread will apply. Their outer expectation and variance turn these varying quantities into numbers.

For a fully explicit mixture distribution, choose a group $Y=A$ or $Y=B$ with equal probabilities. In group $A$, choose $X=1$ or $X=3$ with equal conditional probabilities; in group $B$, choose $X=5$ or $X=11$ with equal conditional probabilities. Thus the complete joint probability distribution consists of $(A,1),(A,3),(B,5),(B,11)$, each with probability $1/4$, and no other outcomes. The conditional means are $2$ and $8$. Squaring the deviations $\pm 1$ and $\pm 3$ gives conditional variances $1$ and $9$, respectively. The law of total expectation gives the overall mean $\mu =(2+8)/2=5$.

The within-group contribution to variance isFor the between-group contribution, the conditional mean takes values $2$ and $8$, each three units from the overall mean. Consequently,We can check this without any conditioning: the discrete random variable $X$ takes values $1,3,5,11$ with equal probabilities, soEquivalently, its second moment is $(1+9+25+121)/4=39$, and subtracting the squared mean gives $39-25=14$. These are population variances, with the stated probabilities as weights, rather than estimates using the corrected denominator of a sample variance.

The figure makes the two sources of spread visible. The outcomes in group $A$ cluster tightly around $2$, while those in group $B$ sit farther apart around $8$. Replacing every outcome by its conditional mean removes the internal spread but leaves the separation between groups. Conversely, subtracting each group's conditional mean aligns the two groups around zero. Averaging only the conditional variances would miss nine units of variance; retaining only the conditional means would miss five. Neither summary alone describes the whole population.

Two modifications clarify what each contribution measures. Shift group $A$ to outcomes $4,6$ and group $B$ to outcomes $2,8$, keeping all four probabilities equal. Both conditional means are now $5$, while the conditional variances remain $1,9$. The between-group contribution vanishes, and total variance is just $5$. Group membership still tells us how uncertain an outcome is, even though the best point prediction stays the same. Alternatively, let every outcome in group $A$ equal $2$ and every outcome in group $B$ equal $8$. Now both conditional variances vanish, and all nine units of total variance come from group separation. Knowing the group predicts the outcome perfectly. These limiting cases distinguish differences in average outcome from differences in internal spread; the original mixture distribution contains both.

The same calculation answers a prediction question. If the group is hidden, the best constant prediction is the overall mean $5$, with mean squared error $14$, by variance as the minimum mean squared error of a constant. If the group is revealed, predict $2$ for $A$ and $8$ for $B$. The errors become $-1,1,-3,3$, and their mean squared error is $5$. Knowing the group therefore improves the best achievable mean squared error by $9$, exactly the between-group contribution. This is an interpretation relative to the information available: the remaining uncertainty could shrink further if additional information were revealed.

For any square-integrable measurable function $a(Y)$ used as a predictor, write $m(Y)=\mathbb{E}[X\mid Y]$. The prediction identity isThus predicting with the regression function $m(y)=\mathbb{E}[X\mid Y=y]$ gives the conditional expectation $m(Y)$ and minimizes mean squared error over every permitted predictor, including nonlinear ones. This minimizes average squared error, not the error for each individual outcome. See Berkeley Statistics 201A notes, sections 36–37.

Why do the two pieces add cleanly? Put $R=X-m(Y)$. Its conditional expectation given $Y$ is zero, so $\mathbb{E}[R]=0$ and $\mathbb{E}[Rb(Y)]=0$ for every square-integrable measurable function $b(Y)$. In particular, $R$ and $m(Y)$ are uncorrelated random variables. Splitting $X-\mu =R+(m(Y)-\mu )$ and squaring therefore leaves no cross term after taking expectation. Geometrically, the conditional expectation is an orthogonal projection in the L2 space, and the law of total variance is its Pythagorean theorem for centered random variables.

In our example, the four products $R(m(Y)-5)$ are $3,-3,-9,9$, which average to zero. Yet $R$ and $Y$ are not independent: observing $|R|=1$ identifies group $A$, while observing $|R|=3$ identifies group $B$. This is a concrete instance of heteroscedasticity, with different conditional variances despite zero conditional means of the errors. A group can predict uncertainty even when it cannot predict the error's sign.

The law of total variance is useful whenever aggregated data conceal distinct groups. Changing mixture weights changes both contributions, so the weighting must match the population being described. For example, giving group $A$ weight $3/4$ changes the overall mean to $7/2$, the within-group variance contribution to $3$, and the between-group contribution to $27/4$, making the total $39/4$. Group differences account for spread and prediction gains; assigning them a causal explanation requires a separate causal inference argument.

A leg of a right triangle is either of the two sides meeting at the right angle. The opposite side is the hypotenuse. For leg lengths $a,b$ and hypotenuse $c$, the Pythagorean theorem gives $a^2+b^2=c^2$, while the area is $ab/2$. In the $3$-$4$-$5$ right triangle, the legs have lengths three and four. Relative to an acute angle, one leg is adjacent and the other opposite; these names depend on the chosen angle.

The modulus of a complex number $z=x+iy$ isIt extends absolute value to the complex plane: the modulus is the Euclidean distance from $z$ to zero, or the Euclidean norm of its coordinate vector. The Pythagorean theorem gives $|3+4i|=5$, even though its two coordinates have sizes $3$ and $4$. The modulus is always nonnegative, and is zero exactly at zero.

Two essential rules areThe second is the triangle inequality. For $z=3+4i$ and $w=1+i$, the product has modulus $\sqrt{50}=5\sqrt{2}\approx 7.071068$, while the sum has modulus $\sqrt{41}\approx 6.403124<5+\sqrt{2}\approx 6.414214$. The strict inequality records that the two displacement vectors point in different directions.

The Euclidean distance between two complex numbers $z,w$ is $|z-w|$. Consequently $|z-w|<\epsilon$ expresses geometric closeness without treating the two coordinates separately. This makes the modulus a natural measure of error in complex analysis; the triangle inequality bounds accumulated error, while the product rule tracks how multiplication changes its size. Equal moduli do not imply equal complex numbers: $5$, $-5$, and $3+4i$ all lie on the same circle around zero. The modulus records size and deliberately forgets direction; the argument supplies the missing information.

The orthogonal projection $P_F$ onto a closed subspace of a Hilbert space $F$ sends $x$ to its component in the orthogonal decompositionThe residual $r$ is orthogonal to every permitted direction. For any $z\in F$, the Pythagorean theorem givesThus $P_{F}x$ uniquely minimizes the distance to $F$. It is a linear map, is idempotent, and is nonexpansive: $\Vert P_{F}x-P_{F}y\Vert \le \Vert x-y\Vert$; taking $y=0$ gives $\Vert P_{F}x\Vert \le \Vert x\Vert$.

In ${\mathbb{R}}^2$, project $v=(1,3)$ onto the line spanned by $u=(2,1)$. The inner product givesThe check $⟨r,u⟩=-2+2=0$ confirms orthogonality, and $\Vert v{\Vert }^2=10=5+5=\Vert P_{F}v{\Vert }^2+\Vert r{\Vert }^2$. Projecting again changes nothing, since the result already lies on the line.

For a finite orthonormal basis $e_1,\dots ,e_m$ of $F$, the same operation is $P_{F}x={\sum }_j⟨x,e_j⟩e_j$, using the convention linear in the first argument. Replacing coordinate vectors by functions gives exactly the same approximation principle. For $f(x)=x^2$ in the real L2 space on $[-1,1]$ and $F=\mathrm{span}\{1,x\}$, the directions are orthogonal, with squared norms $2$ and $2/3$. Their inner products with $f$ are $2/3$ and zero, soDirect integration gives the squared error $\Vert x^2-1/3{\Vert }^2=8/45$. For any competing affine polynomial, the orthogonal residual yieldsEvery deviation from $a=1/3$, $b=0$ adds a nonnegative error. The best function need not pass through any selected data point: it minimizes an integrated error across the whole interval.

If $(u-p,q)=0$ for all $q\in A$, then for any $a\in A$ the difference $p-a$ lies in $A$ and Pythagorean theorem givesThus $p$ is best.

Conversely, if $p$ is best, then for every $q\in A$ the real quadratichas its minimum at $t=0$. Its linear coefficient must vanish, giving $(u-p,q)=0$. Hence best approximation is equivalent to orthogonality of the residual to the approximation space. The weight is positive almost everywhere, so this is a genuine positive-definite inner product on continuous functions.

On $V$, the bilinear form $a$ is symmetric, and part (b) proves $a(v,v)\ge \alpha \Vert v{\Vert }_1^2$. Thus $a$ is an inner product: $a(v,v)\ge 0$, and equality forces $v=0$. Its induced energy norm is $\Vert v{\Vert }_a=\sqrt{a(v,v)}$. For any real scalar $t$,The Cauchy-Schwarz inequality for this inner product follows directly by expanding $a(p-tq,p-tq)\ge 0$ and minimizing over $t$ when $q\ne 0$. Hencewhich proves the triangle inequality. This verifies every norm axiom. In fact, the estimates from part (b) give the equivalent norms

Subtract the discrete weak formulation from the continuous one to get Galerkin orthogonality,For any $v_h\in V_h$, write $u-v_h=(u-u_h)+(u_h-v_h)$. The second summand is in $V_h$, so expansion and Galerkin orthogonality give the Pythagorean theorem in the energy norm:The last term is nonnegative and vanishes exactly when $v_h=u_h$. Choosing $v_h=u_h$ attains the lower bound, provingThus $u_h$ is the unique best approximation; the map $u\mapsto u_h$ is the orthogonal projector onto $V_h$ for the linear elasticity energy norm.