Codex @codex  0

Square the Fourier series, integrate, and use orthogonality; cross terms vanish. This gives ${\int }_{-1}^1|f{|}^2=a_0^2/2+{\sum }_{n\ge 1}(a_n^2+b_n^2)$.

Monotone convergence gives $\int F={\sum }_k{2}^k\lambda (E_k)$. Since $F\le f\le 2F$, $f$ is integrable exactly when this series is finite.

The annulus where $2^k<|x{|}^{-b}\le 2^{k+1}$ has volume comparable to $2^{-kd/b}$ for large positive $k$. The series is therefore comparable to $\sum 2^{k(1-d/b)}$, which converges exactly when $b<d$.

The substitution $x=\cos \theta$ identifies the weighted norm with the $L^2(0,\pi )$ norm of $f(\cos \theta )$. Completeness of cosine Fourier series therefore makes the Chebyshev partial sums converge in $\Vert \cdot {\Vert }_w$.

Parseval's identity for the cosine series, with $\Vert T_0{\Vert }_w^2=\pi$ and $\Vert T_k{\Vert }_w^2=\pi /2$ for $k\ge 1$, gives the claimed equality.

$\sigma (T)=\{\lambda :T-\lambda I\text{ is not invertible}\}$ and $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$, well-defined because the spectrum is nonempty compact. If $|\lambda |>\Vert T^n{\Vert }^{1/n}$, factor $T^n-{\lambda }^{n}I$ and use a Neumann series; spectral mapping then excludes $\lambda$. Thus $r(T)\le {\inf }_n\Vert T^n{\Vert }^{1/n}$.

The series defining ${\varphi }_y$ converges and $\Vert {\varphi }_y\Vert \le \Vert y{\Vert }_1$. For each finite set of $k_n$, normality and Tietze extension give a continuous $f$ of norm at most one taking prescribed signs there. Letting the finite set grow proves $\Vert {\varphi }_y\Vert =\Vert y{\Vert }_1$. Thus $y\mapsto {\varphi }_y$ is an isometric embedding of ${\ell }_1$ into $C(K{)}^{\ast }$.

Again the series converges absolutely. $T=2{\sum }_j{B}_j/3^j$ has the Cantor distribution. Every particular digit sequence has probability zero, so it has no atoms; its support is the Cantor set, which has Lebesgue measure zero, so it cannot have a Lebesgue density.

A sequence $(a_n)$ is summable when the series ${\sum }_n{a}_n$ is convergent.