THESIS WALKTHROUGH · SUBSECTION 1.2.2 · LEMMA 1.18

Why Legendre modes stay within one

From a point on the unit circle to an input on the interval, then to the value of a polynomial.

\(\displaystyle \sup_{x\in[-1,1]}|P_n(x)|=1,\qquad n\in\mathbb N_0.\)

Next: ellipse growth · planned lesson

Input x = cos t

Polynomial value Pₙ(x)

Sum of weights

One conjugate pair determines the input

On the unit circle, taking a reciprocal reflects the point across the real axis. Their imaginary parts cancel at the midpoint. What remains is their common horizontal coordinate.

\begin{aligned}z=e^{it},\quad z^{-1}=e^{-it}=\overline z,\\ x=\mathit{Jouk}(z)=\frac{z+\overline z}{2}=\cos t.\end{aligned}

The gold midpoint is the input. The purple mean is the polynomial value at that input. They agree for \(n=1\), but generally differ.

Try n = 2 and t = 90°: the input is 0 while the polynomial value is −1/2.

The polynomial is a balance point of unit phases

Put a positive mass at each phase point on the circle. The mass is its structural weight. The purple point is the balance point of all these masses, so it lies inside their convex hull.

P_n(\cos t)=\sum_{j=0}^n\alpha_{n,j}e^{i(n-2j)t},\qquad\alpha_{n,j}=\alpha_j\alpha_{n-j}>0.

\sum_{j=0}^n\alpha_{n,j}=P_n(1)=1.

The weights sum to one because at \(t=0\) every phase is \(1\), and the Legendre normalization is \(P_n(1)=1\). Opposite frequencies have equal weights. Their imaginary parts cancel, leaving the complete mean on the real axis.

The disk gives an upper bound. An endpoint attains it.

The unit disk is convex. A weighted mean with non-negative weights summing to one cannot leave it. Algebraically, this is the triangle inequality:

\begin{aligned}|P_n(\cos t)|&\le\sum_{j=0}^n\alpha_{n,j}|e^{i(n-2j)t}|\\&=\sum_{j=0}^n\alpha_{n,j}=1.\end{aligned}

Every input in \([-1,1]\) is \(\cos t\) for some \(t\in[0,\pi]\). This bounds the whole interval. Press Align all phases: at \(t=0\), they all point to \(1\), so \(P_n(1)=1\).

\sup_{x\in[-1,1]}|P_n(x)|\le1\quad\hbox{and}\quad|P_n(1)|=1.

The argument proves the supremum. The sampled curve illustrates it.

More oscillations, the same absolute area

Multiplication by \(P_n(\cos t)\) cannot make the height exceed \(|\sin(kt)|\). The signed area can cancel, so its magnitude is at most the area under the absolute envelope.

\sigma_{n,k}=\int_0^\pi P_n(\cos t)\sin(kt)\,dt,\qquad k\in\mathbb N.

\begin{aligned}|\sigma_{n,k}|&\le\int_0^\pi|P_n(\cos t)|\,|\sin(kt)|\,dt\\&\le\int_0^\pi|\sin(kt)|\,dt=2.\end{aligned}

There are \(k\) absolute sine lobes. Each has width \(\pi/k\) and area \(2/k\), so their combined area is \(k\cdot(2/k)=2\).

Input plane · why cos t?

\(e^{it}\) and \(e^{-it}\) have midpoint \(x=\cos t\).

● Point● Conjugate● Input midpoint

Output plane · the weighted mean

All phase points have modulus \(1\). Their weights need not be equal.

◎ Selected phase◌ Its partner● Mean Pₙ(x)

The same input and value on the real graph

The moving point is \((x,P_n(x))\). The dashed levels are \(y=\pm1\).

Inspect the weights and phase frequencies

\(\alpha_\ell=\binom{2\ell}{\ell}4^{-\ell}\). Choose a row to highlight that phase and its reciprocal partner.

Step 1 of 4

Source, assumptions and the next subsection

This lesson follows Subsection 1.2.2 — Real-line stability of Legendre modes, in Section 1.2, Legendre coefficient bounds and projection tails on Bernstein ellipses.

Lemma 1.18 — Real-line sup bound. For every \(n\in\mathbb N_0\),

\sup_{x\in[-1,1]}|P_n(x)|=1.

The thesis attributes this result to Bandtlow’s EDMD–Legendre Lemma 1.2 and standard Legendre normalization. That is the cited source’s numbering; Lemma 1.18 is its number in the thesis.

We use the unnormalised Legendre polynomials, radius one, \(0\le t\le\pi\), and positive integer \(k\). The limiting trace is the interval \([-1,1]\); it is not a nondegenerate Hardy ellipse.

The next subsection studies growth on genuine ellipses with \(\varrho>1\). This later lesson is not available yet. The later nonnegative parity ladder for the kernel coefficients is not established by the examples here.