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The relevant fixed coordinate subspace is $\{(u,0):u\in {\mathbb{R}}^k\}$. Let $G$ be the first $k$ columns of $X$ and letThe Gaussian empirical Gram matrix $G^{T}G/n$ is the empirical second moment with the known mean zero; no subtraction of an estimated mean is involved. For $\theta =(u,0)$, the ratio under consideration is $|u^{T}Du|/\Vert u{\Vert }_2^2$. Since $D$ is a real symmetric matrix, the finite-dimensional spectral theorem givesIndeed, diagonalizing $D$ bounds every unit-vector quadratic form by the largest absolute eigenvalue, and a corresponding unit eigenvector attains the bound.

We first construct a unit sphere net from ball covering. Enlarge the numerical covering constant, if necessary, to $A_0\ge 1$. Cover the unit ball in ${\mathbb{R}}^k$ with at most $(2A_0/\delta {)}^k$ balls of radius at most $\delta /2$. For each such ball meeting the unit sphere, choose a point of the sphere in it and discard the others. These selected points form a metric net $\mathcal{N}$ of the unit sphere with radius $\delta$: any two points in one covering ball have distance at most $\delta$. This argument ensures that the net points have unit length even if the original covering centers did not.

Take $\delta =1/4$ and put $B=8A_0$, so $|\mathcal{N}|\le B^k$. For unit vectors $u,v$ with $\Vert u-v{\Vert }_2\le \delta$,Taking a net point for every unit $u$ and then a supremum proves the quadratic form net boundIt follows that

Fix $v\in \mathcal{N}$. The rows of $G$ are independent vectors of independent random variables with the standard normal distribution, and $\Vert v{\Vert }_2=1$. Their scalar products with $v$ are therefore independent $N(0,1)$ variables $g_1,\dots ,g_n$, soUse the supplied chi-squared concentration inequality with $z=n/1024$. Its threshold isConsequently $P(|v^{T}Dv|>1/4)\le 2e^{-n/1024}$. The union bound gives the stronger fixed-subspace estimateThis is Gaussian Gram matrix concentration on a fixed subspace.

Since $1\le k<p$, we have $p\ge 2$. Under $n\ge Ck\log p$,Choose the numerical constant $C\ge 1024(1+\log B/\log 2)$. Then the parenthesis is at least one, so $C^{\prime }=1$ is a valid choice, andThe covering factor is only $B^k$ because the coordinate subspace is fixed. The ambient dimension enters through the assumed sample-size bound.

The original PDF has $u_m^{n+1}$ on the left of the update; the converted TeX omits its $+1$. Use the Forward Euler method with $k=\Delta t\ge 0$ and $\mu =k/h_x^2$. The explicit time stepping for bounded reaction diffusion update matrix isWhen $0\le \mu \le 1/2$, the heat update $H_h$ has nonnegative stencil weights $\mu ,1-2\mu ,\mu$ and row sums at most one. Hence its induced maximum operator norm is at most one. It is also a symmetric matrix with eigenvaluesall in $[-1,1]$, so its discrete L2 norm operator bound is at most one too. Let $A_{\ast }=\max (|a_0|,|a_{+}|)$. In either of these norms,Iteration gives the mesh-uniform stability estimateThe same estimate applies to differences of solutions. Adding a per-step defect $k{r}^n$ gives $\Vert e^n\Vert \le e^{A_{\ast }T}(\Vert e^0\Vert +k{\sum }_{j<n}\Vert r^j\Vert )$.

This proof only requires $\mu \le 1/2$ and bounded reaction coefficients. It does not assume the full update $E_h$ is nonnegative: a negative $a_m$ can make its middle stencil weight negative when $\mu =1/2$. Nor does it claim contractivity or an all-time bound for every coefficient. The required stability is a bound uniform in the discretization on fixed finite intervals.

A complex square matrix $U$ is a unitary matrix when $U^{†}U=I$, equivalently $U^{-1}=U^{†}$. Here $†$ denotes the adjoint matrix, the conjugate transpose. Such a matrix preserves the standard complex inner product and its Euclidean norm.

A real symmetric matrix $A$ is a Hermitian matrix, so $A^{†}=A$. For arbitrary complex $\mathbf{x}$,The imaginary cross terms cancel because $A$ is Hermitian. Reversing the signs gives the same identity:In particular, $(A+iI)\mathbf{x}=0$ implies $|\mathbf{x}{|}^2=0$, so $A+iI$ has zero kernel of a linear map. It is a square matrix, and the rank-nullity theorem therefore makes it invertible. The same argument proves invertibility of $A-iI$.

The proposed matrixis consequently well-defined. For any $\mathbf{y}$, put $\mathbf{x}=(A+iI{)}^{-1}\mathbf{y}$. The two norm identities show directly thatTo verify the unitary matrix identity explicitly, take adjoints:The two shifts commute, because both are polynomials in $A$. HenceThusThe imaginary shifts are invertible, and their equal norm changes cancel in the quotient. This is the Cayley transform of a Hermitian matrix in the $(A-iI)(A+iI{)}^{-1}$ convention.

Use completing the square rather than computing the eigenvalues of the symmetric matrix. Successively removing the mixed terms givesThus one suitable invertible linear change of coordinates isThe coordinate matrix is upper triangular with determinant one, so the change is an invertible linear map. Its inverse is $z=Z$, $y=Y+4Z/7$, $x=X-2Y+5Z/14$. Hence the diagonal coefficients can be taken as $2,-7,-3/14$. This is a matrix congruence of a quadratic form; an orthogonal transformation was not required.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelöf theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

Assume $u\in W^{1,2}(B_2(0))$ is nonnegative almost everywhere, and that the real coefficient matrix $A=(a^{ij})$ is measurable and satisfies, for almost every $x$ and every $\xi \in {\mathbb{R}}^n$,The operator norm bound controls all coefficients, including any antisymmetric part. For a symmetric matrix, these assumptions are exactly the usual lower and upper quadratic-form bounds. The weak solution condition isThen the Harnack inequality for uniformly elliptic divergence-form equations isA nonnegative weak solution has its supremum controlled by its infimum on the smaller ball. The De Giorgi-Nash-Moser theorem supplies a Hölder continuous function representative, for which ordinary supremum and infimum can replace the essential supremum and essential infimum. Nonnegativity and the larger ball are essential hypotheses; there is no differentiability requirement on the coefficients.

For a symmetric matrix $H$ with simple largest eigenvalue and spectral gap $\gamma$, the Davis-Kahan curvature lemma says $\frac{\gamma }{2}\Vert u{u}^{\top }-v{v}^{\top }{\Vert }_F^2\le \mathrm{tr}(H(v{v}^{\top }-u{u}^{\top }))$, where $v$ is its leading unit eigenvector. Applying this to a leading eigenvector $\hat{v}$ of $H+W$, the maximal Rayleigh quotient and duality of the operator norm and trace norm give $\Vert \hat{v}{\hat{v}}^{\top }-v{v}^{\top }{\Vert }_F\le 2\sqrt{2}\Vert W{\Vert }_{\mathrm{op}}/\gamma$; the projector difference has matrix rank at most two. This is a rank-one eigenprojector perturbation bound.

Here $W=M-M_0=A-A_0$ has independent centered bounded upper-triangular entries, apart from symmetry. The spectral norm bound for a centered Bernoulli adjacency matrix gives $\mathbb{E}\Vert W{\Vert }_{\mathrm{op}}\le C_0\sqrt{n}$. Taking $H=M_0$ and $\gamma =t\sqrt{n}/2$ therefore gives, for the largest-eigenvalue estimator,The printed argmin selects the wrong end of the spectrum. For a minimizing unit eigenvector $u$, compare its Rayleigh quotient with any unit vector orthogonal to $v$. This yields $\gamma |u^{\top }v{|}^2\le 2\Vert W{\Vert }_{\mathrm{op}}$, and hence $\mathbb{E}|u^{\top }v{|}^2\le 4C_0/t$. Its mean projector error is at least $\sqrt{2}(1-4C_0/t)$, using $\sqrt{1-x}\ge 1-x$. Along $t=n^{1/4}$ with sufficiently large even $n$, the printed parameter restriction holds and this lower bound tends to $\sqrt{2}$, contradicting an error of order $1/t$. Replace argmin by argmax, or negate the entire matrix before minimizing.

Over $\mathbb{R}$ or $\mathbb{C}$, a Lie algebra is a vector space with a bilinear map $[,]$ which is alternating and obeys the Jacobi identity:Bilinearity and alternation imply $[X,Y]=-[Y,X]$.

For a Matrix Lie group, identify the tangent space $\mathfrak{g}=T_{I}G$ with derivatives ${\gamma }^{\prime }(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real vector space, even when the matrices have complex entries. For $X,Y\in \mathfrak{g}$, choose a curve $\gamma (s)$ with ${\gamma }^{\prime }(0)=X$. Conjugating a curve with derivative $Y$ shows thatDifferentiate this curve in the finite-dimensional vector space $\mathfrak{g}$. The result isThe matrix commutator is bilinear and alternating, and expanding the six terms proves its Jacobi identity. Thus this construction gives the Lie algebra of a matrix Lie group, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.

For the unitary group, differentiating $\gamma (t{)}^{†}\gamma (t)=I$ gives $X^{†}+X=0$. Conversely, if $X^{†}=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. ConsequentlyThis is the unitary Lie algebra. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A matrix-unit basis of the unitary Lie algebra isThese $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$ anti-Hermitian matrix units and their combinations are linearly independent over $\mathbb{R}$ and span every allowed entry.

The symplectic stabilizer is a subgroup: the identity preserves $J$, and if $MJ{M}^T=J$ and $NJ{N}^T=J$, thenMultiplying $MJ{M}^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}J{M}^{-T}=J$. Products and inverses remain unitary. This is the compact symplectic group, often denoted $USp(2n)$ or $Sp(n)$.

Differentiating the stabilizer equation gives $XJ+J{X}^T=0$. For $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, this saysTogether with $X^{†}=-X$, these are equivalentlyThese conditions are also sufficient: $e^{tX}$ is unitary, andHence it stays in the subgroup. They characterize its compact symplectic Lie algebra without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex symmetric matrix $B$ has $n(n+1)$ real parameters. Thus

Here is a complete matrix-unit basis of the compact symplectic Lie algebra. The generators supplying $A$ areThe two families supplying the real and imaginary parts of $B$ are, for $1\le i\le j\le n$,The denominator merely avoids double-counting diagonal entries. Each displayed generator obeys both defining tangent conditions. The $H,P,Q$ generators form a real basis of the allowed $A$ blocks, and the $U,V$ generators form a real basis of the complex symmetric $B$ blocks. Thus they are independent and their total number is $n(2n+1)$. They give all generators required by the real compact algebra, including the case $n=1$.

The real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

Use the natural cubic spline convention appropriate to the interior design knots: the spline is linear on $[a,x_1]$ and $[x_n,b]$. We also need $n\ge 2$. These points will be important for uniqueness.

A cubic spline is a function in $C^2[a,b]$ whose restriction to each of $[a,x_1]$, $[x_1,x_2]$, through $[x_n,b]$ is a polynomial of degree at most three. A spline knot is a junction between consecutive pieces. For a natural cubic spline, the two exterior pieces are affine functions; equivalently, one uses the natural endpoint conditions $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching derivatives. Merely prescribing $g^{\prime \prime }(a)=g^{\prime \prime }(b)=0$ on otherwise arbitrary exterior cubic pieces is a different condition and does not provide the unique interpolant invoked in this question.

The natural cubic spline interpolant $N\mathbf{v}$ to values $\mathbf{v}=(v_1,\dots ,v_n{)}^T$ is the natural cubic spline satisfying $(N\mathbf{v})(x_i)=v_i$. Its existence and uniqueness for $n\ge 2$ are the interpolant fact allowed in the PDF. The interpolation map $N$ is a linear map: a linear combination of interpolants satisfies the same spline, boundary and value conditions, so uniqueness identifies it with the interpolant to the corresponding linear combination of value vectors.

Let $g=N\mathbf{v}$, let $\tilde{g}\in C^2[a,b]$ have the same values, and put $r=\tilde{g}-g$. Then $r(x_i)=0$. We prove the orthogonalityApply integration by parts twice on each polynomial interval. On an interior interval, $g^{\prime \prime \prime \prime }=0$, andThe $g^{\prime \prime \prime }r$ terms vanish because $r$ vanishes at the knots. The $g^{\prime \prime }{r}^{\prime }$ terms cancel at internal knots because $g^{\prime \prime }$ and $r^{\prime }$ are continuous, and vanish at $x_1,x_n$ because $g^{\prime \prime }$ is zero there. On the exterior intervals, $g^{\prime \prime }=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g^{\prime \prime \prime }$ is continuous across knots.