Chebyshev polynomials are orthogonal for the arcsine weight

By [F1]–[F2], $⟨T_m,T_n{⟩}_C={\int }_0^{\pi }\cos (m\theta )\cos (n\theta )d\theta$. If $m\mathrm{\ne }n$, [F3], cosine parity [F5], and $\sin (k\pi )=0$ from [F5] make the integral of each nonzero-frequency cosine vanish by [F4]. If $m=n\ge 1$, [F3] gives $\frac{1}{2}(1+\cos (2m\theta ))$, whose integral is $\pi \mathrm{/}2$. If $m=n=0$, the integrand is $1$ and its integral is $\pi$. Linearity [F10] passes from the product-to-sum identity to these integrals.

The recurrence [F6] shows inductively that $T_n$ has degree $n$ and nonzero leading coefficient: this holds for $T_0,T_1$, and the leading term of $2X{T}_n$ cannot cancel with the lower-degree $T_{n-1}$. Eliminating the leading term of any polynomial successively proves that $T_0,T_1,\dots$ span all polynomials; their distinct degrees also give linear independence. Step 1.1 gives positive squared norms and pairwise orthogonality.

Suppose continuous $h$ pairs to zero with every $T_n$. By step 2.1 it pairs to zero with every polynomial. Fix $\epsilon >0$ and choose a polynomial $q$ with $\mathrm{\parallel }q-h{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ by [F7]. The bound $M=\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ follows from [F8]. The pointwise bounds $-M\epsilon \le h(\cos \theta )(h-q)(\cos \theta )\le M\epsilon$ and [F9] give $0\le ⟨h,h{⟩}_C=⟨h,h-q{⟩}_C\le \pi M\epsilon$, where the equality uses bilinearity [F10]. The lower inequality uses the nonnegative integrand $h(\cos \theta {)}^2$ and [F9]. Since this holds for every $\epsilon >0$, $⟨h,h{⟩}_C=0$. If $h(x_0)\mathrm{\ne }0$ at some $x_0\in [-1,1]$, [F11] supplies ${\theta }_0\in [0,\pi ]$ with $\cos {\theta }_0=x_0$; continuity makes $h(\cos \theta {)}^2$ bounded below by a positive constant on a one-sided neighborhood if ${\theta }_0$ is an endpoint, or a two-sided neighborhood otherwise. [F9] then gives a positive integral, contradicting zero norm. Therefore $h=0$.

For any continuous $h$, if $⟨h,h{⟩}_C=0$ but $h(\cos {\theta }_0)\mathrm{\ne }0$ at some ${\theta }_0\in [0,\pi ]$, continuity gives a one-sided interval at an endpoint, or a two-sided interval in the interior, on which $h(\cos \theta {)}^2$ is bounded below by a positive number. [F9] makes its integral positive, a contradiction. Thus $h(\cos \theta )=0$ for every $\theta$, and [F11] gives $h=0$ on $[-1,1]$. Symmetry and bilinearity follow from the integral definition [F1] and linearity [F10], proving positive definiteness.