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Use the weighted inner productMultiply the equation for $R_n$ by ${\overline{R}}_m$, subtract the complex conjugate equation for $R_m$ times $R_n$, and integrate. Since the eigenvalues are real,The boundary term vanishes as before. Thus eigenfunctions with distinct eigenvalues are orthogonal. The full disk inner product differs only by the angular factor $2\pi$.

Use the ordinary real inner product $⟨u,v⟩={\int }_a^{b}uvdx$ on the domain of functions obeying the two separated boundary conditions. Integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $a$, the two boundary-value vectors $(u(a),u^{\prime }(a))$ and $(v(a),v^{\prime }(a))$ satisfy the same homogeneous linear relation. If $A_2=0$, both function values vanish; if $A_2\ne 0$, their derivatives are the same constant multiple $-A_1/A_2$ of their values. Either way $v{u}^{\prime }-u{v}^{\prime }=0$. The same argument applies at $b$. HenceThis proves symmetry, and the regular separated Sturm-Liouville problem has the same adjoint domain, so $L$ is self-adjoint. To see the domain point, require the boundary form to vanish against every admissible $u$: the independently variable endpoint data force an adjoint-domain function to obey those same endpoint relations.

As usual, each endpoint pair $(A_1,A_2)$ and $(B_1,B_2)$ must be nonzero, and the coefficients must have the regularity needed for these operations. An identically zero pair would impose no condition at that endpoint and would not give the claimed self-adjoint operator. The related operator ${\omega }^{-1}L$ is self-adjoint in the weighted inner product ${\int }_a^b\omega uvdx$.

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

Let $A=A^{†}$ be a Hermitian matrix and $Av=\lambda v$ with $v\ne 0$. The scalar $v^{†}Av$ equals its own complex conjugate, becauseSince $v^{†}Av=\lambda v^{†}v$ and the squared norm $v^{†}v$ is positive, $\boxed{\lambda \in \mathbb{R}}$.

Now take $Au=\lambda u$ and $Av=\mu v$. The inner product satisfiesSubtracting gives $(\mu -\lambda )u^{†}v=0$. Thus distinct eigenvalues imply $\boxed{u^{†}v=0}$, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.

A complex square matrix $U$ is a unitary matrix if $\boxed{U^{†}U=U{U}^{†}=I}$. Equivalently, it preserves the complex inner product and hence the norm. If $Uv=\mu v$ with $v\ne 0$, thenDivision by the positive squared norm gives $\boxed{|\mu |=1}$. Thus every eigenvalue lies on the unit circle.

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

Project the initial condition onto the orthogonal spatial eigenfunctions. Their squared norms in the ordinary inner product areThe latter follows from the integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ being $1/2$, with the sine-cosine cross term zero. ThereforeFor square-integrable data the expansion converges to the data in the L2 norm at $t=0$; for $t>0$ the decaying high modes are smoothed. Initial data with zero $e^x$ projection have no growing term.

Apply separation of variables with $\rho (x,t)=X(x)T(t)$ and temporal factor $T=e^{-\lambda t}$. The spatial eigenvalue equation isThe nontrivial solutions of these Dirichlet boundary conditions are sine eigenfunctions. Normalizing in the usual inner product on $(0,L)$ giveswith ${\int }_0^L{\psi }_n{\psi }_{m}dx={\delta }_{nm}$. The superposition principle and the Fourier sine series then giveThe constants are the initial-data coefficients $a_n={\int }_0^L\rho (x,0){\psi }_n(x)dx$. Here physical diffusion means $D>0$.

For sufficiently smooth functions satisfying the same boundary conditions, integration by parts gives the boundary formThe stated endpoint relation annihilates this expression. This proves the required self-adjoint differential operator identity; for the usual regular Sturm-Liouville problem, separated real boundary conditions give a self-adjoint realization on its domain.

In this problem the transformed diffusive flux is $J=-e^{-\lambda t}p{y}^{\prime }$. Thus the physical boundary conditions areequivalently $y^{\prime }(\pm L)=0$ when the endpoint values of $p$ are positive. In particular $p{f}^{\prime }=p{g}^{\prime }=0$ at the endpoints, which makes the boundary form vanish directly. The inner product used here is the ordinary integral; the weight $w$ enters the generalized eigenvalue problem, not this boundary-form calculation.

Multiplication by $x^2$ gives the Sturm-Liouville formso the weight is $w(x)=x^2$. Use the weighted inner product $⟨f,g{⟩}_w={\int }_0^{\pi /2}{x}^2\overline{f}gdx$ and the Sturm-Liouville operator $L=-x^{-2}(x^{2}d/dx{)}^{\prime }$. Integration by parts gives its boundary form:The upper endpoint term vanishes by the Dirichlet boundary conditions, and the lower one vanishes for the regular solutions. More precisely, the radial Dirichlet spectrum in three dimensions identifies $L$ under the unitary transformation $v=xu$ with $-v^{\prime \prime }$ on $(0,\pi /2)$ with zero endpoint values; this specifies the self-adjoint realization at the singular endpoint as well as proving the formal identity.

For eigenfunctions $u_a,u_b$ with different eigenvalues, the same identity gives $({\lambda }_b-{\lambda }_a)⟨u_a,u_b{⟩}_w=0$. Taking real eigenfunctions proves the required orthogonality of Sturm-Liouville eigenfunctions.

The norm induced by the specified inner product is the L2 normA polynomial $p_n\in {\Pi }_n$ is a best polynomial least-squares approximation whenMinimizing this norm is equivalent to minimizing its square, the integrated squared error. Continuity ensures that a function with zero squared norm is identically zero: a nonzero value would give a neighbourhood on which the square has a positive integral.

Projection onto the constant polynomials requires the residual to be orthogonal to $1$. For the original inner product,For the weighted inner product the corresponding normal equation isHenceIt increases monotonically towards one. Although the weight vanishes at zero, it is positive elsewhere, and a nonzero continuous function is nonzero on an interval of positive weight. Thus the weighted expression really is an inner product and the constant minimizer is unique.

The normalized weight $(2n+1)x^{2n}/2$ has total mass one and assigns mass $r^{2n+1}$ to $[-r,r]$ for each $0<r<1$. Its mass therefore concentrates near $-1$ and $1$, where $x^2=1$. This explains why the weighted best constant tends to one rather than to the unweighted spatial average. It is endpoint concentration in weighted constant approximation.

The orthonormal columns satisfy $\Vert a{\Vert }_2=\Vert b{\Vert }_2=1$ and $a^{T}b=0$. By part (i), $Q_{1}a=e_1$. A product of orthogonal matrices is an orthogonal matrix, since $Q_1^T{Q}_1=I$. ConsequentlyThe last expression is the first coordinate of the transformed second column. Thus the first row isThe one in the first entry uses the nonnegative retained-coordinate convention. A different consistent sign convention could give $(-1,0)$ instead; the second entry must still be zero by preservation of the inner product.

Write ${\mathcal{P}}_n$ for the vector space of polynomials of polynomial degree at most $n$, and put $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$. As the question calls the weighted integral an inner product, its finiteness is implicit; for example, a positive integrable weight $w$ suffices for all the continuous functions involved here.

For every $q\in {\mathcal{P}}_n$, differentiating the squared weighted norm $\Vert f-p_n-tq{\Vert }_w^2$ at its minimum $t=0$ gives the orthogonality equationsIn particular, the requested linear system isIts Gram matrix $G_{kj}={\int }_{-1}^{1}w(x)x^{k+j}dx$ is a symmetric positive-definite matrix: for any nonzero vector $a$,Indeed a nonzero polynomial is nonzero on some open interval, and the weight is positive. Therefore $G$ is an invertible matrix, proving uniqueness. Conversely, if these equations hold, the Pythagorean theorem in an inner-product space gives, for every $q\in {\mathcal{P}}_n$,so the solution really is the unique best approximation.

The power iteration starts from $v_0\ne 0$, normally scaled to unit Euclidean norm, and repeatsNormalization prevents the rapidly growing or decaying magnitudes of $A^k{v}_0$ from obscuring its direction. The iteration requires $w_{k+1}\ne 0$.

Because $A$ is a diagonalizable matrix, choose an eigenbasis $u_1,\dots ,u_n$, with corresponding eigenvalues ${\lambda }_1=\lambda ,{\lambda }_2,\dots ,{\lambda }_n$, and expand $v_0={\sum }_j{c}_j{u}_j$. First suppose the dominant eigenvalue is simple and $c_1\ne 0$. ThenForthe sum in parentheses after $u_1$ is $O(q^k)$ in Big O notation. Consequently the power iteration direction error is $O(q^k)$: the iterates approach the line spanned by $u_1$. For a real matrix, a uniquely dominant eigenvalue is real, since otherwise its complex conjugate would have the same modulus. If $\lambda >0$, $v_k$ approaches one unit eigenvector; if $\lambda <0$, successive $v_k$ alternate signs. In either case,This is the sign and phase in power iteration qualification. Also, the necessary assumption is a nonzero coefficient $c_1$ in the eigenbasis, rather than merely a nonzero ordinary inner product with $u_1$: those tests can differ for a nonsymmetric matrix. If $c_1=0$, the method cannot generate the missing component.

If the dominant value $\lambda$ has a higher-dimensional eigenspace, the same argument replaces $c_1{u}_1$ by the nonzero sum of all starting components in that eigenspace. The limiting direction is then that of this sum, which is still an eigenvector for $\lambda$. The usual algorithm assumes $\lambda \ne 0$; a strict dominance gap guarantees this whenever another eigenvalue is present.

An estimate retaining the sign of $\lambda$ is the Rayleigh quotientContinuity of this quotient at a nonzero eigenvector proves the limit even when $A$ is a nonsymmetric matrix. Alternatively, $(A{v}_k{)}_i/(v_k{)}_i\to \lambda$ for a coordinate whose limiting eigenvector component is nonzero. The scaling factor $\Vert A{v}_k{\Vert }_2$ alone tends only to $|\lambda |$.

The Gauss-Seidel method updates coordinates successively using the newest available entries:It is well-defined for every initial vector exactly when every $a_{ii}\ne 0$, equivalently when the triangular matrix $D+L$ is invertible. Nonsingularity of $A$ alone does not ensure this. The matrix form isIf the iterates converge to $z$, passing to the limit gives $(D+L)z=b-Uz$, hence $Az=b$. Nonsingularity identifies $z$ as the unique solution $x$.

Now assume $A$ is a symmetric positive-definite matrix. Testing with the coordinate vector gives $a_{ii}=e_i^{T}A{e}_i>0$, so $D$ is positive definite. The form $⟨v,w{⟩}_A=v^{T}Aw$ is an inner product: bilinearity and symmetry are immediate and positivity is the hypothesis. Its Cauchy-Schwarz inequality givesproving the triangle inequality; definiteness and absolute homogeneity are immediate. Thus $\Vert v{\Vert }_A=\sqrt{v^{T}Av}$ is the energy norm.

Subtract the iteration from the exact equation to obtain $(D+L)e_{k+1}=-L^T{e}_k$. HenceUsing the preceding iteration also gives $(D+L)e_k=-L^T{e}_{k-1}$ and therefore $A{e}_k=L^T(e_k-e_{k-1})$ for $k\ge 1$, as required.

Put $d=e_k-e_{k+1}$. Expanding the two quadratic forms and using the first identity gives the exact Gauss-Seidel energy decrease identity:Here $2d^{T}Ld=d^T(L+L^T)d$. If $d=0$, then $A{e}_k=0$, so $e_k=0$. Otherwise $d^{T}Dd>0$. Thus $\boxed{\Vert e_k{\Vert }_A>\Vert e_{k+1}{\Vert }_A}$ unless the exact solution has already been reached.

Let $f_i=f(ih)$ and $g_i=g(ih)$, and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions $U_0=V_0=A$ and $U_N=V_N=B$. These equations determine the interior nodal values.

For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.

For any $q$ in the polynomial trial space, $f-q=(f-p)+(p-q)$ and $p-q$ is in that same space. The assumed orthogonality therefore gives the Pythagorean identityThus $p$ is the best approximation; equality is possible only when $q=p$ because the inner product is positive-definite.

Write the linear polynomial as $p=A+Bx$. Orthogonality of $x^2-p$ to $1$ and $x$ in the weighted inner product givesSolving yieldsThe weighted squared error is ${\int }_0^{1}x(x^2-6x/5+3/10{)}^{2}dx=1/600$, providing an additional exact check. The weight's zero value at one endpoint does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval of positive weight.

The form is symmetric and bilinear by the corresponding properties of finite real sums. Its squared norm is $⟨p,p⟩={\sum }_{i}p(x_i{)}^2\ge 0$. Equality means $p$ vanishes at all $n+1$ distinct nodes. A nonzero polynomial of degree at most $n$ has at most $n$ roots, so equality forces $p=0$. Thus this is an inner product on ${\Pi }_n$.

On ${\Pi }_{n+1}$ it is only positive semidefinite: the nonzero node polynomial $\omega (x)={\prod }_{i=0}^n(x-x_i)$ satisfies $⟨\omega ,\omega ⟩=0$. Hence it is not an inner product on ${\Pi }_{n+1}$. The finite-evaluation inner product on polynomials works exactly when the evaluation map is injective on the chosen space.

Finally, if $q\ne 0$, positivity applied to $p-tq$ and the choice $t=⟨p,q⟩/⟨q,q⟩$ giveRearranging proves Cauchy-Schwarz inequality in its required form:If $q=0$ both sides of the asserted inequality are zero. Equality holds exactly when $p,q$ are linearly dependent, since in the nonzero case the minimizing $p-tq$ must have zero norm.

For any $q\in {\Pi }_n$, write $f-q=(f-p)+(p-q)$. The stated orthogonality makes the two summands perpendicular in the given inner product, because $p-q\in {\Pi }_n$. The Pythagorean theorem for inner product spaces therefore givesEquality is attained at $q=p$, and positive definiteness makes it unique. Thus $p$ is the orthogonal projection of $f$ onto the polynomial subspace.

For the weighted example set $p(x)=A+Bx$. Orthogonality against the basis $1,x$ givesOdd terms vanish on the symmetric interval, so $A=3/5$ and $B=0$. ConsequentlyThe weight $x^2$ vanishes only at one point; its integral against $g^2$ is positive for every nonzero continuous $g$, so it really does define an inner product. For an optional error check, the squared minimum error is $2/7-12/25+6/25=8/175$.