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Suppose a linear map between finite-dimensional Hilbert spaces has matrix $A$, and their inner products in the chosen bases are $⟨x,z{⟩}_H=z^{\ast }{G}_{H}x$ and $⟨u,v{⟩}_K=v^{\ast }{G}_{K}u$. The Gram matrices $G_H,G_K$ are positive-definite matrices. Writing $A^{\ast }={\overline{A}}^{\mathrm{T}}$ for the ordinary conjugate transpose, the coordinate matrix of the adjoint operator isIndeed, the defining inner product identity says $y^{\ast }{G}_{K}Ax=y^{\ast }[T^{\ast }{]}^{\ast }{G}_{H}x$ for all $x,y$. Thus simply transposing and conjugating is valid in orthonormal bases, where both Gram matrices are the identity matrix.

For the weighted inner product $⟨x,y{⟩}_G=\overline{y_1}{x}_1+4\overline{y_2}{x}_2$ and the coordinate swap $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, the adjoint operator has matrix $\left(\begin{array}{cc}0& 4\\ 1/4& 0\end{array}\right)$. With $x=(1,0{)}^{\mathrm{T}}$ and $y=(0,1{)}^{\mathrm{T}}$, both inner products equal $4$. The coordinate swap is its own ordinary transpose, but is not self-adjoint in this weighted inner product.

Let $Ay=(p{y}^{\prime }{)}^{\prime }+qy$ have real eigenfunctions $u_n$ obeying $A{u}_n={\lambda }_{n}w{u}_n$ and homogeneous Dirichlet boundary conditions, with $N_n={\int }_a^{b}w{u}_n^2>0$. Suppose $Ay+\mu wy=f$, with possibly nonzero endpoint values. The Lagrange identity for a Sturm-Liouville operator givesDivision by $({\lambda }_n+\mu )N_n$ gives the displayed coefficient whenever the denominator is nonzero. At resonance in a differential equation, equality of the forcing projection and the boundary term is the Fredholm solvability condition. The Sturm-Liouville eigenfunction expansion represents the interior solution in the weighted inner product sense; its zero endpoint terms do not reproduce nonzero boundary traces by pointwise substitution. A smooth lift of the endpoint data reduces the calculation to homogeneous Dirichlet boundary conditions.

For $0<x<1$ and zero endpoint values, write $L=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ with $p=x^{-2}$, $w=x^2$ and $q=-a{x}^2$. The change of variable $z=x^3/3$ has $dz=x^{2}dx$ and sends $y^{\prime }=x^2{Y}_z$, $y^{\prime \prime }=2x{Y}_z+x^4{Y}_{zz}$. Thus it is an isometry from the weighted inner product to the ordinary integral on $0<z<1/3$, and the apparent singular operator becomes $-Y_{zz}-aY$. Its Dirichlet eigenvalues are $9n^2{\pi }^2-a$ and its normalized eigenfunctions are $\sqrt{6}\sin (n\pi x^3)$, $n\ge 1$. The singular endpoint is handled through this transformed regular domain; a zero eigenvalue occurs exactly when $a=9n^2{\pi }^2$.

Let $f$ be a real function with $\Vert f{\Vert }_w^2={\int }_a^{b}f(x{)}^{2}w(x)dx<\infty$, and let $V_m$ be the vector space of real polynomials of degree at most $m$. Its orthogonal projection is the least-squares polynomial in an orthogonal-polynomial basisEach coefficient depends on just one weighted inner product with $f$ and one squared norm. In particular, the coefficients $c_0,\dots ,c_m$ stay the same when the allowed polynomial degree increases. This statement assumes a fixed weight, interval and normalization of the orthogonal polynomials.

To see why the least-squares normal equations decouple, first write a trial polynomial as $q={\sum }_{j=0}^m{a}_j{\varphi }_j$ in any basis of $V_m$. The residual must be orthogonal to each permitted direction, givingThe Gram matrix usually couples the unknown coefficients. With ${\varphi }_j=p_j$, it becomes $G_{ij}=h_i{\delta }_{ij}$, leaving the separate equations $h_i{a}_i=b_i$. For the monomial basis on $[-1,1]$ with $w=1$, by contrast, $G_{ij}$ is zero when $i+j$ is odd and $2/(i+j+1)$ when $i+j$ is even. Distinct even powers, or distinct odd powers, remain coupled.

There is also a proof of the minimum that uses no differentiation. The defining coefficients give $⟨f-q_m,p_n{⟩}_w=0$ for $0\le n\le m$. Thus $f-q_m$ is orthogonal to all of $V_m$. For every $v\in V_m$, the Pythagorean theorem in an inner-product space givesThe second term is nonnegative and vanishes exactly when $v=q_m$, proving that the least-squares solution is unique. This is an integrated squared-error optimum; it does not assert a smallest error at every point or a smallest uniform norm.

For $N>m$, the nested orthogonal projections satisfy the exact residual budgetIndeed, $q_N-q_m$ is a linear combination of mutually orthogonal polynomials, and $f-q_N$ is orthogonal to it. Every newly retained component removes exactly $c_n^2{h}_n$ from the squared error. No infinite expansion or completeness assumption is needed for this finite identity. The Legendre least-squares projections of a cubic make every term explicit.

The Chebyshev differential equation $-(1-x^2)y^{\prime \prime }+x{y}^{\prime }=\lambda y$ on $-1<x<1$, with zero endpoint values, has the Sturm-Liouville formThus $p=\sqrt{1-x^2}$ and the weight defining the weighted inner product is $w=(1-x^2{)}^{-1/2}$. The substitution $x=\cos \theta$ maps the problem to $-Y^{\prime \prime }=\lambda Y$ on $0<\theta <\pi$ with homogeneous Dirichlet boundary conditions. It also transforms $w(x)dx$ into $-d\theta$. Consequently the normalized eigenfunctions areTheir weighted inner product is the ordinary sine inner product on $[0,\pi ]$. Equivalently, $y_n=\sqrt{2/\pi }\sqrt{1-x^2}{U}_{n-1}(x)$, where $U_{n-1}$ is a Chebyshev polynomial of the second kind. These are not the cosine solutions $T_n$, which have different endpoint values. The original endpoints are singular because $p$ vanishes and $w$ diverges there, but the transformed Dirichlet problem specifies the self-adjoint realization. The eigenfunctions vanish at both endpoints and their quasi-derivatives $p{y}_n^{\prime }$ have finite limits, so the boundary form vanishes.

For $-y^{\prime \prime }=\lambda x^{-2}y$ on $1<x<e^{\ell }$ with zero endpoint values, the natural weighted inner product has measure $dx/x^2$. Set $v=\ln x$ and $y(x)=\sqrt{x}Z(v)$. The chain rule gives $x^2{y}^{\prime \prime }=\sqrt{x}(Z^{\prime \prime }-Z/4)$, so $-Z^{\prime \prime }+Z/4=\lambda Z$ with zero endpoints on $(0,\ell )$. Moreover $|y{|}^{2}dx/x^2=|Z{|}^{2}dv$, so the transformation preserves the norm. The ordinary sine orthonormal eigenbasis therefore gives exactly the displayed spectrum and eigenfunctions for $n\ge 1$. The factor $\sqrt{x}$ and the additive $1/4$ distinguish this problem from the operator $-x^2{y}^{\prime \prime }-x{y}^{\prime }$ with measure $dx/x$.

Let $f$ be a real continuous function on a nondegenerate compact interval and let $p_n$ be its best approximation among polynomials of polynomial degree at most $n$ for a positive integrable weighted inner product. Then $f-p_n$ has at least $n+1$ distinct zeros. Indeed, if it had at most $n$, multiply it by the polynomial with simple zeros at precisely its interior sign changes. The product has constant nonzero sign away from finitely many zeros, so its weighted integral is nonzero, contradicting orthogonality of $f-p_n$ to all polynomials of polynomial degree at most $n$.

On real polynomials with the weighted inner productthe differential operator $Lf=x{f}^{\prime \prime }+(1-x)f^{\prime }$ is symmetric. Indeed, $e^{-x}Lf=(x{e}^{-x}{f}^{\prime }{)}^{\prime }$ and integration by parts givesThe factor $x$ removes the boundary term at zero and the exponential removes it at infinity. On the finite-dimensional polynomial vector space of degree at most $n$, $L$ preserves the space and is a self-adjoint operator. Its eigenvalues are $0,-1,\dots ,-n$, with the ordinary Laguerre polynomials as eigenvectors. On the full weighted $L^2$ Hilbert space, symmetry on the polynomial domain alone does not assert equality of adjoint operator domains; a self-adjoint realization requires its operator domain to be specified.

Take the polynomial $f(x)=1+x+x^2+x^3$ on $[-1,1]$ with unit weight. Using the four Legendre polynomials above, direct integration givesFor example, $⟨f,P_2⟩=\frac{1}{2}{\int }_{-1}^1(3x^4+2x^2-1)dx=4/15$: the omitted odd terms have zero integral. Dividing by $h_2=2/5$ gives $c_2=2/3$. Substituting all four coefficients proves the exact identity

The successive orthogonal projections onto degrees at most $m$ are thereforeHere $\Vert g{\Vert }_2^2={\int }_{-1}^{1}g(x{)}^{2}dx$ is the squared L2 norm, without division by the interval length. Each row retains the previous Legendre polynomial coefficients and appends one new term. The ordinary power coefficients can change: adding $\frac{2}{3}{P}_2$ changes the constant from $4/3$ to $1$, and adding $\frac{2}{5}{P}_3$ changes the linear coefficient from $8/5$ to $1$. Independence belongs to the chosen orthogonal polynomial coordinates.

The full squared L2 norm is $\Vert f{\Vert }_2^2=192/35$, equal to $32/9+128/75+8/45+8/175$. The residual budgets are the omitted terms:For a direct check independent of summing the budgets, $f-q_2=x^3-3x/5$, and integrating its square gives $2/7-12/25+6/25=8/175$. This residual is orthogonal to $1,x,x^2$, exactly the conditions for the best quadratic least-squares solution. An orthogonal projection need not be a polynomial interpolant: for example, $q_2(1)=18/5$ while $f(1)=4$. The approximation allocates error across the interval according to the chosen weighted inner product.

For a Cauchy-Euler equation, logarithmic coordinates turn scale changes in $x>0$ into translations in $t$. With $Y(t)=y(e^t)$, the chain rule gives $x{y}^{\prime }=Y^{\prime }$ and $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Consequently the Sturm-Liouville operator $-x^2{y}^{\prime \prime }-x{y}^{\prime }-cy$ becomes $-Y^{\prime \prime }-cY$. Its natural weighted inner product has measure $dx/x=dt$, so this change of variable also preserves the norm when the transformed problem uses ordinary integration in $t$. On $1<x<e^{\ell }$ with Dirichlet boundary conditions, the orthonormal eigenfunctions are $\sqrt{2/\ell }\sin (n\pi \log x/\ell )$ and the eigenvalues are $(n\pi /\ell {)}^2-c$, for $n\ge 1$.

For $x>0$, write $t=\log x$ and $y=xv(t)$. The chain rule gives $y^{\prime }=v+v_t$, $y^{\prime \prime }=x^{-1}(v_t+v_{tt})$, and thereforeOn $1<x<e^{\ell }$ with homogeneous Dirichlet boundary conditions, the eigenfunctions are $x\sin (n\pi \log x/\ell )$ and the eigenvalues are $-(n\pi /\ell {)}^2$. The weighted inner product with weight $x^{-3}$ becomes ${\int }_0^{\ell }\overline{v}wdt$, so their squared norms are $\ell /2$ and their orthogonality is exactly that of the Fourier sine series. The weight is the factor that makes this conjugacy preserve the inner product.

With the weighted inner product ${\int }_{-1}^1\overline{f}gdx/\sqrt{1-x^2}$, the Chebyshev differential equation has the Sturm-Liouville form with coefficients $p=\sqrt{1-x^2}$, $w=(1-x^2{)}^{-1/2}$ and $q=0$. The substitution $x=\cos \theta$ converts its operator to $-d^2/d{\theta }^2$ on $(0,\pi )$ and its inner product to the ordinary one in $d\theta$. The natural endpoint conditions $p{y}^{\prime }\to 0$ become zero derivative conditions at $\theta =0,\pi$.

Consequently the eigenfunctions are the Chebyshev polynomials $T_n(x)=\cos (n\arccos x)$ with eigenvalues $n^2$, $n=0,1,2,\dots$. Their squared norms are $\pi$ for $n=0$ and $\pi /2$ otherwise. In particular $T_1=x$ and $T_3=4x^3-3x$ are orthogonal. The endpoints of the original differential expression are singular; specifying these transformed Neumann boundary conditions defines its self-adjoint realization. The Dirichlet eigenfunctions of the Chebyshev differential operator use a different endpoint condition and have sine rather than cosine profiles.

The Laguerre polynomial normalized by $L_n(0)=1$ has coefficients $L_n(x)={\sum }_{j=0}^n(-1{)}^j(\genfrac{}{}{0ex}{}{n}{j})x^j/j!$. By the Leibniz rule, this is equivalent to Rodrigues' formula, $e^{-x}{L}_n(x)=D^n(e^{-x}{x}^n)/n!$. For any polynomial $q$ of degree below $n$, $n$ applications of integration by parts give ${\int }_0^{\infty }{e}^{-x}{L}_{n}qdx=0$. All endpoint terms vanish: derivatives of $e^{-x}{x}^n$ of order below $n$ vanish at zero and decay at infinity. Since $L_n^{(n)}=(-1{)}^n$, the same calculation givesThus these orthogonal polynomials are already an orthonormal set for the weighted inner product with weight $e^{-x}$.

Let $L=w^{-1}[-(d/dx)(pd/dx)+q]$ be the regular self-adjoint operator of a Sturm-Liouville problem, with eigenfunctions orthonormal for the weighted inner product $\int w\overline{u}v$. For $\mu$ outside its spectrum, the displayed kernel satisfies $(L_x-\mu )G_{\mu }=\delta (x-\xi )$ with an ordinary, unweighted Dirac delta. Indeed weighted completeness gives $w(\xi ){\sum }_n{y}_n(x)y_n(\xi {)}^{\ast }=\delta (x-\xi )$. Thus $u(x)=\int G_{\mu }(x,\xi )f(\xi )d\xi$ solves $(L-\mu )u=f$. A kernel integrated against $w(\xi )d\xi$ instead omits the prefactor. For real coefficients, the weighted reciprocity relation is $G_{\mu }(x,\xi )/w(\xi )=G_{\mu }(\xi ,x)/w(x)$ when the spectral parameter is real and outside the spectrum. A zero eigenvalue prevents inversion at $\mu =0$ without a forcing compatibility condition or a projected inverse.

An orthogonal polynomial family chooses polynomials that do not overlap under a specified inner product. For real polynomials on an interval $(a,b)$, use the weighted inner productAssume $a<b$, the weight $w$ is measurable and positive almost everywhere, and its absolute moments ${\int }_a^b|x{|}^{k}w(x)dx$ are finite for every integer $k\ge 0$. These conditions make the integrals finite and give every nonzero polynomial positive squared norm. A sequence $p_0,p_1,\dots$ is an orthogonal polynomial family when $p_n$ has real coefficients and polynomial degree exactly $n$, and $⟨p_n,p_k{⟩}_w=0$ for $n\ne k$.^[ref]

Here $h_n=\Vert p_n{\Vert }_w^2>0$ and ${\delta }_{nk}$ is the Kronecker delta. Orthogonality does not require $h_n=1$: an orthonormal set is obtained only after replacing $p_n$ by $p_n/\sqrt{h_n}$. Multiplying each member by any nonzero real constant preserves orthogonality, but changes its norm and the corresponding expansion coefficient. The monic orthogonal polynomial convention instead fixes each leading coefficient to one.

The first four Legendre polynomials, named for Adrien-Marie Legendre, use $w=1$ on $[-1,1]$ and the normalization $P_n(1)=1$:Their orthogonality and squared norms areThus their Gram matrix has a positive diagonal and zero entries elsewhere, although their function graphs cross and overlap on the page. Orthogonality measures an integral of products, rather than separation of function graphs.^[ref]

For these four polynomials, a short verification expands each product and uses ${\int }_{-1}^1{x}^{2j+1}dx=0$ and ${\int }_{-1}^1{x}^{2j}dx=2/(2j+1)$. For example, $⟨P_0,P_2⟩=(3(2/3)-2)/2=0$ and $⟨P_1,P_3⟩=(5(2/5)-3(2/3))/2=0$. The same calculation gives the diagonal entries $2,2/3,2/5,2/7$. More generally, apply the Gram-Schmidt process to the monomial basis $1,x,x^2,\dots$: at step $n$, subtract projections onto the earlier polynomials. The result retains its leading term, has degree $n$, and has positive norm, proving existence without a differential equation. Different degrees also give linear independence, so $p_0,\dots ,p_m$ form a basis of the polynomials of degree at most $m$.

This basis turns least-squares solutions for functions into independent scalar calculations, as the decoupled coefficients of orthogonal polynomial projections show. In a spectral method, increasing the degree enlarges the approximation space while keeping earlier projection coefficients fixed. Weighted integrals also connect the same orthogonal polynomials to Gaussian quadrature: its nodes are polynomial roots of a member of the family. The projection problem minimizes an integrated squared error; Gaussian quadrature uses the family to choose evaluation points for an integral.

Two vectors are orthogonal when their inner product is zero.

For real Euclidean vectors, the inner product is the dot product, so $(3,4)$ and $(4,-3)$ are orthogonal: $3(4)+4(-3)=0$. For nonzero Euclidean vectors, this means their angle is $90^{\circ }$. The zero vector is orthogonal to every vector, although it has no direction or defined angle.^[ref]

In an abstract inner product space, orthogonality always refers to its chosen inner product. For example, the weighted inner product $⟨u,v{⟩}_w=4u_1{v}_1+u_2{v}_2$ makes $(1,1)$ and $(1,-4)$ orthogonal, since $4-4=0$. Their ordinary dot product is $-3$, so their arrows are not perpendicular in the usual Cartesian coordinate system. The algebraic definition also applies to functions, where arrows need not be available.

For a positive even weight on a symmetric interval, any orthogonal polynomial of degree $k$ has the parity of $k$. Induct on $k$. The reflected polynomial $q_k(x)=(-1{)}^k{p}_k(-x)$ has the same leading coefficient. By substitution $x\mapsto -x$ in the weighted inner product and the inductive parity of all lower polynomials, it is orthogonal to them. Hence $q_k-p_k$ has lower degree and is orthogonal to the entire space containing it. Its squared norm is zero, forcing $q_k=p_k$. The base case is a constant. This proof needs neither monic nor unit-norm normalization; it only compares polynomials with the same leading coefficient.

Put $x=\cos \theta$ in the weighted inner product. Since $T_r(\cos \theta )=\cos (r\theta )$,The last equality follows by the product-to-sum trigonometric identity and integration. Thus these are nonzero mutually orthogonal polynomials. The first $n+1$ are linearly independent, and the vector space $P_n$ has dimension $n+1$; therefore $\boxed{P_n=\mathrm{span}(T_0,\dots ,T_n)}$.

For any $q\in P_n$, the function of the real parameter $t$has a minimum at $t=0$. Its derivative there vanishes, proving $\boxed{(u-p_n,q)=0}$. This is precisely the defining orthogonality of an orthogonal projection, or a least-squares polynomial in an orthogonal-polynomial basis. Conversely, this condition and the Pythagorean theorem for inner product spaces give $\Vert u-q{\Vert }_2^2=\Vert u-p_n{\Vert }_2^2+\Vert p_n-q{\Vert }_2^2$, proving the minimum and its uniqueness.

Expand $p_n={\sum }_{r=0}^n{a}_r{T}_r$ in the basis already established and take the inner product with $T_s$. Orthogonality leaves $a_s(T_s,T_s)=(u,T_s)$, soIn particular $a_0={\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )d\theta$ and $a_r=2{\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )\cos (r\theta )d\theta$ for $r\ge 1$. These are the corresponding Chebyshev series coefficients with the constant coefficient convention used in this question.

Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is $c_k=N_k^{-1}{\int }_1^{e}ry{\hat{y}}_{k}dx$. Although ${\hat{y}}_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\hat{L}y=r{e}^x-(\beta -1)ry$. Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on $\alpha$.

If $\beta =1+n^2{\pi }^2$ for a positive integer $n$, the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of ${\hat{y}}_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.

Finally, for $\alpha \ne 0$, the displayed Sturm-Liouville eigenfunction expansion represents the solution on $1<x<e$ and in the weighted inner product sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand $y-h$ with homogeneous Dirichlet boundary conditions and add $h$ back afterwards.

Match the coefficient of $y^{\prime }$ with the derivative of the coefficient of $y^{\prime \prime }$. For Sturm-Liouville form this requiresA positive integrating factor is therefore $r(x)=x^{-3}$, with any positive constant multiple equally valid. Direct calculation givesThis uses the plus-divergence sign convention of the question. The usual minus-divergence Sturm-Liouville operator is its negative; no extra sign is inserted into the requested eigenvalue equation.

Because $r$ never vanishes on the interval, multiplying the original eigenvalue equation by $r$ leaves the requested weighted eigenvalue problem unchanged:The weight in the inner product is essential. The substitution $t=\log x$ givesThus orthogonality of Sturm-Liouville eigenfunctions holds for the weighted inner product with weight $x^{-3}$, and the squared norm is $N_k=1/2$.