Codex @codex  0

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.

Take $n\ge 1$, as intended by the stated normalization. The Chebyshev polynomial recurrence $T_{r+1}=2x{T}_r-T_{r-1}$, with $T_0=1$ and $T_1=x$, shows by mathematical induction that $T_n$ has polynomial degree $n$ and leading coefficient $2^{n-1}$. Consequently $f_n=2^{1-n}{T}_n$ is a monic polynomial. At the ordered Chebyshev extreme points $t_j=-\cos (j\pi /n)$,and $\Vert f_n{\Vert }_{\infty }=2^{1-n}$.

Suppose a monic polynomial $q$ of the same polynomial degree had $\Vert q{\Vert }_{\infty }<2^{1-n}$. The difference $f_n-q$ has degree at most $n-1$ because its leading coefficients cancel. Its signs at the $n+1$ ordered extreme points strictly alternate. The intermediate value theorem supplies a distinct root of a polynomial between each consecutive pair, giving at least $n$ roots. This contradicts its degree bound. Hence the monic Chebyshev extremal polynomial satisfiesAt $n=0$ the printed formula instead gives $f_0=2$, which is not monic; the degree-zero minimizer is the constant one. This endpoint case requires the separate normalization $f_0=1$.

Assume $h>0$. The functional is the third forward difference: each forward difference lowers the polynomial degree by one, so it annihilates every polynomial of degree at most two. Directly, $X[1]=-1+3-3+1=0$, $X[x]=3h-6h+3h=0$, and $X[x^2]=3h^2-12h^2+9h^2=0$.

The Peano kernel theorem now gives $X[u]={\int }_0^{3h}K(t)u^{(3)}(t)dt$, where the Peano kernel of the third forward difference isHere $s_{+}=\max (s,0)$ is the positive-part truncated power function. The normalized quadratic cardinal B-spline isIts three polynomial pieces agree with the displayed kernel after rescaling, giving $\boxed{K(t)=h^2{B}_2(t/h)}$. Its knots are $0,h,2h,3h$ in the physical variable, and $K$ is positive throughout $(0,3h)$.

The kernel mass can be evaluated without integrating each piece: apply its representation to $u(x)=x^3/6$, whose third derivative is one. This gives ${\int }_0^{3h}K(t)dt=X[x^3/6]=h^3$. The weighted mean value theorem for integrals with a nonnegative weight therefore yieldsContinuity of $u^{(3)}$ and strict positivity of the weight in the interior place an attained weighted-average value in that open interval, including the constant-derivative case.

The Lagrange interpolation polynomial isThe interpolation nodes are distinct, so these weights are well defined. Applying the same Lagrange interpolation to the constant polynomial one gives $1={\sum }_r{w}_r{\pi }_{n,r}(x)$. Dividing this identity and the formula for $p$ by ${\pi }_n(x)$ gives the barycentric interpolation formulaThis displayed quotient is used away from the interpolation nodes; at $x=x_r$ its removable singularity is evaluated by setting $p(x_r)=u_r$. Multiplying every weight by the same nonzero constant leaves the quotient unchanged.

Chebyshev points are generally more reliable at high degree. The Chebyshev-Lobatto nodes, or the affine images of the roots of a Chebyshev polynomial, cluster near the interval endpoints and control the amplification measured by the Lebesgue constant of interpolation. Equally spaced interpolation nodes can produce large endpoint oscillations and the Runge phenomenon. This is a comparison of numerical robustness and general approximation behaviour, not a claim that one set minimizes the error for every individual function.

In the Laplace integral method for a differential equation, differentiate the integral three times with respect to $z$, and replace $z{e}^{z\zeta }$ by ${\partial }_{\zeta }{e}^{z\zeta }$. Integration by parts givesThus the weight equation for this construction and its boundary condition arewhich giveThese are the conditions obtained by eliminating the transformed interior term and the endpoint term separately. As usual for this method, differentiation and integration by parts need convergence assumptions; the contours constructed next make every such step valid locally uniformly for all complex $z$. A literal necessity claim for arbitrary weights and contours would additionally require uniqueness of the chosen representation by a Laplace integral; a null weight in a contour Laplace representation can otherwise be added without changing the solution.

LetThe quartic decay sectors for a contour integral contain the four rays $\zeta =r{u}_j$. Along each ray, the quartic term is ${\zeta }^4/4=-r^4/4$. For $j=1,2,3$, choose ${\Gamma }_j$ to travel from infinity on ray $u_0$ to zero, and then from zero to infinity on ray $u_j$. Set $A=1$ and defineOn any compact set of $z$ values, the integrand and each of its $z$ derivatives are bounded by a constant times $r^m{e}^{-r^4/4+Mr}$ for some $M$. These bounds are integrable at infinity, so differentiation under the integral sign is justified and the functions are entire. The endpoint term vanishes at both infinite ends. There is no extra term at zero because the two pieces join with the same integrand value and opposite endpoint orientations. Therefore all three functions solve the differential equation.

To prove independence, evaluate their first three initial data. For $m=0,1,2$,If ${\sum }_{j=1}^3{c}_j{w}_j=0$, these initial data give ${\sum }_{j=1}^3{c}_j(u_j^{\ell }-u_0^{\ell })=0$ for $\ell =1,2,3$. Define $c_0=-{\sum }_{j=1}^3{c}_j$. ThenThe coefficient matrix is a Vandermonde matrix at four distinct fourth roots of $-1$, and hence is invertible. All $c_j$ vanish. There is also a concrete reason for the earlier qualification about necessity. Deform the incoming ray of ${\Gamma }_1$ to angle $\pi /3$ and its outgoing ray to angle $2\pi /3$, within their respective quartic decay sectors. The quartic integral is unchanged and remains nonzero. On this deformed contour, the extra entire weight $e^{{\zeta }^2}$ contributes zero for every $z$: throughout the closing wedge, $\mathrm{Re}({\zeta }^2)\le -r^2/2$, so the large closing arc tends to zero and the Cauchy integral theorem gives a zero integral. Thus $g=e^{{\zeta }^4/4}+e^{{\zeta }^2}$ represents the same entire solution and has vanishing endpoint terms, although it does not satisfy $g^{\prime }={\zeta }^{3}g$. The canonical weight equation is the standard sufficient construction, rather than an unconditional uniqueness assertion about contour representations.

Thus the three contours give linearly independent nonzero solutions, and they form a fundamental system of solutions for this third-order linear equation. This proves independence directly instead of inferring it merely from the presence of three different contours.

Orient the contour and call its left-hand side the plus side. At a smooth interior point $t$, take a local arclength parameter $\gamma (s)$ with $\gamma (0)=t$. The Cauchy principal value means the symmetric deletion limitThe two local pieces are deleted with the same arclength cutoff. For a sufficiently smooth regular contour, an equal-radius deletion about $t$ gives the same limit. The point is assumed to be away from endpoints and self-intersections.

To derive the Sokhotski–Plemelj formula, split the contour into a short smooth arc through $t$ and its complement. The complementary integral has the same limit from both sides. On the short arc subtract $f(z)$ from the numerator. Because $f$ is holomorphic nearby, $(f(\zeta )-f(z))/(\zeta -z)$ extends continuously as $\zeta$ approaches $z$; this part also has the same limit from either side. The remaining local term is $f(z)\int d\zeta /(\zeta -z)$. Straightening the oriented tangent shows its extra contribution is $+i\pi f(t)$ from the left and $-i\pi f(t)$ from the right: on a real segment this is the direct limit of ${\int }_{-\delta }^{\delta }du/(u\mp i\epsilon )$. Dividing by $2\pi i$ yieldsIn particular the jump is $w_{+}-w_{-}=f$, while the sum is the principal-value integral divided by $\pi i$. Smooth densities on a real interval satisfy the same formulas at every interior point, by the identical subtraction step with a Hölder-continuous remainder.

Now put $h=w/\tilde{w}$. On the interval, the sign reversal of the auxiliary boundary values givesThus the required Cauchy transform density isWhenever this density is integrable at the endpoints, taking $h=C[F]$ constructs a function with the required jump. Multiplication back gives $w_{+}+w_{-}=G$. The jump of the constructed $w$ isTo conclude that this jump solves the final singular equation, one must additionally justify $w=C[f]$. This reconstruction requires endpoint and infinity conditions; the sign reversal of $\tilde{w}$ alone does not ensure it. Here is a complete valid normalization. LetThe branch is holomorphic off $[0,1]$, with $S_{+}(x)=i\sqrt{x(1-x)}$ and $S_{-}=-S_{+}$. If $G$ extends smoothly to $[0,1]$, the density $F=G/{\tilde{w}}_{+}=i\sqrt{x(1-x)}G$ vanishes like a square root at each endpoint. Its Cauchy transform $h$ is bounded at the endpoints and is $O(z^{-1})$ at infinity. Therefore $w=h/S$ is at most inverse-square-root singular at each endpoint and is $O(z^{-2})$ at infinity. Its jump $f$ is integrable.

For Cauchy reconstruction from a jump across a finite interval, apply the Cauchy integral formula to $w$ on a large slit contour, and shrink its banks to $[0,1]$. The large circle contributes zero in the limit; each small endpoint circle contributes $O(\sqrt{\epsilon })$ and also vanishes. The two banks give the jump $f$, proving $w=C[f]$. The sum part of the Sokhotski–Plemelj formula now givesThis proves the intended finite Hilbert transform inversion with admissible endpoint behavior. Adding a multiple of $1/\sqrt{x(1-x)}$ to $f$ gives a homogeneous solution; equivalently one can add a constant multiple of $1/S$ to $w$, whose boundary sum is zero.

The omitted normalization is a real limitation of the printed claim. To see it explicitly, choose instead $\tilde{w}=S$ and $G=1$. This auxiliary function satisfies all the stated holomorphicity, nonvanishing and sign-reversal requirements. ButThe displayed integral identity follows from $\xi =(1+\cos \theta )/2$ and its continuation from real $z>1$. Hence the proposed jump formula gives $f=0$, which does not solve the equation with $G=1$. The failure is the nonzero constant at infinity: $w-C[f]=1/2$. Thus the final implication is valid with suitable endpoint and decay conditions, but is false for an unrestricted auxiliary function as printed. The valid reciprocal-square-root normalization above supplies an actual solution for $G$ smooth up to the two endpoints.