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For uniform convergence to $k$ on $A$, for every $\eta >0$ there is $N$ such that $|k_n(x)-k(x)|<\eta$ for all $n\ge N$ and all $x\in A$. The bound $|\sin (nx)/\sqrt{n}|\le 1/\sqrt{n}$ proves uniform convergence to zero, whereas its derivative is $\sqrt{n}\cos (nx)$. On an interval containing zero these derivatives diverge at zero, so they cannot converge uniformly. This supplies a counterexample to differentiating a uniformly convergent sequence without additional hypotheses.

For uniform derivative convergence with an anchored value, put $g=\lim k_n^{\prime }$ and $a=\lim k_n(x_0)$. A uniform limit of continuous functions is continuous. The fundamental theorem of calculus givesThus $k_n(x)\to k(x)=a+{\int }_{x_0}^{x}g(t)dt$, with error bounded byThis is uniform on bounded subintervals, and on all of $A$ if it is bounded. The fundamental theorem of calculus now gives $k^{\prime }=g$, so $k$ is continuously differentiable. For an unbounded $A$, no global uniform convergence of $k_n$ follows merely from the displayed hypotheses, but local uniform convergence and the claimed differentiability do.

For the series, use partial sums and differentiate their individual terms:For $|x|\le 1$, their absolute values are at most $2n/(n^3+1)$, a summable bound. The Weierstrass M-test makes the derivative series uniformly convergent, while every partial sum is zero at $x=0$. The anchored-value result therefore proves continuous differentiability on $(-1,1)$, with derivative equal to the displayed termwise derivative series.

The original PDF has indices $3^k$; the exponent is lost in the TeX transcription. This lacunary indexing is essential to the positive lacunary Chebyshev series argument below.

Let $K$ be the least nonnegative integer with $3^K>n$, so $K=0$ for $n=0$. DefineFor $K=0$ the sum is empty and means the zero polynomial. Each included Chebyshev polynomial has degree at most $n$. Also $|T_j(x)|\le 1$ on the interval, so summability of the positive coefficients gives uniform convergence by the Weierstrass M-test, andChoose $L=3^K$ and the $L+1$ points $x_j=\cos (j\pi /L)$, $0\le j\le L$. For every omitted index $k\ge K$, the integer $3^{k-K}$ is odd. Consequently,Every term of the tail has the same sign at a given point, and thereforeThis shows both that the error norm is exactly $A_K$ and that it alternates at $L+1\ge n+2$ distinct points. The points are in decreasing order; reversing their order still gives alternation. The Chebyshev alternation theorem proves that this partial sum is the unique best uniform approximation. HenceIn particular, $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$. The equal signs of all tail terms at the same extrema are the reason positivity and the odd integer frequency ratios are useful.

An explicit construction gives Bernstein's lethargy theorem in the requested inequality form. SetStrict decrease makes every coefficient positive. Telescoping and the limit assumption giveThusdefines a continuous function by the Weierstrass M-test and the uniform limit theorem. Apply the positive lacunary Chebyshev series calculation. At $n=0$,For $n\ge 1$, choose $K\ge 1$ so that $3^{K-1}\le n<3^K$. Its error isTherefore the function satisfiesThis explicit Chebyshev construction for Bernstein lethargy also has $E_n(f_{ϵ})\to 0$. It shows that continuity imposes no universal speed of convergence of best polynomial approximation, even though convergence itself follows from the Weierstrass approximation theorem.

By the Weierstrass M-test, absolute summability of the Fourier coefficients makes ${\sum }_n\hat{f}(n)e^{int}$ uniformly convergent to a continuous periodic function $H$. Termwise integration is justified by that uniform convergence, and orthogonality givesPart (ii) then implies $H=f$. Thus the Fourier series converges uniformly to the original function, not merely to some continuous limit. In particular,

For $0<x<1$, $|x^k/k^2|\le 1/k^2$, and the numerical series ${\sum }_{k\ge 1}{k}^{-2}$ converges. The Weierstrass M-test therefore gives uniform convergence to $f(x)={\sum }_{k\ge 1}{x}^k/k^2$. More explicitly,The bound is independent of $x$, including points arbitrarily close to one.

The Chebyshev polynomial identity $T_m(\cos \theta )=\cos (m\theta )$ implies $|T_m(x)|\le 1$ on $[-1,1]$. Since ${\sum }_i{\gamma }_i<\infty$, the Weierstrass M-test shows thatconverges uniformly on $[-1,1]$. Every partial sum is continuous, so the uniform limit theorem makes $f$ a well-defined continuous function.

LetThen $-1=x_0<x_1<\cdots <x_{3^{n+1}}=1$. Since every $3^i$ is odd and, for $i\ge n+1$, $3^{i-n-1}$ is an odd integer,It follows thatThus the error has equal alternating extrema at $3^{n+1}+1$ ordered points. The Chebyshev alternation theorem shows that $P_n$ is a best uniform approximation among polynomials of degree at most $3^{n+1}-1$, and in particular

Now let $({\delta }_N)$ be the prescribed decreasing positive sequence. DefineThen $s_j$ decreases strictly to zero, so every ${\gamma }_j>0$, the series ${\sum }_j{\gamma }_j=s_0$ converges, andGiven $N\ge 1$, choose $j\ge 0$ with $3^j\le N\le 3^{j+1}-1$. Monotonicity of best-approximation errors and of ${\delta }_N$ givesHence this continuous $f$ satisfiesfor every polynomial $P$ of degree at most $N$. This is the polynomial form of Bernstein's lethargy theorem.

For the period lattice $\Lambda =⟨\lambda ,\mu ⟩$, define the Weierstrass elliptic function byTo prove Normal convergence of the Weierstrass elliptic-function series, let $K\subset \mathbb{C}\setminus \Lambda$ be compact and choose $R$ with $|z|\le R$ on $K$. For $|\omega |>2R$,The comparison series ${\sum }_{\omega \ne 0}|\omega {|}^{-3}$ converges by the stated lattice-sum criterion. The Weierstrass M-test gives uniform convergence on $K$; since $K$ was arbitrary, the series converges locally uniformly away from $\Lambda$. The locally uniform convergence of holomorphic functions makes its sum holomorphic there, and the displayed principal term gives a double pole at every lattice point.

The function $\wp$ is even and satisfiesThe three nonzero two-torsion points $z_1,z_2,z_3$ are distinct modulo $\Lambda$ and each equals its own negative modulo $\Lambda$. Therefore no two can have the same $\wp$-value, so $e_1,e_2,e_3$ are distinct. Equivalently, $\wp$ is a degree-two branched map and each half-lattice point already occupies its fiber with multiplicity two.

For a transitive example, take the equianharmonic latticeand define $\theta ([z])=[\omega z]$. Multiplication by $\omega$ preserves $\Lambda$, so $\theta :\mathbb{C}/\Lambda \to \mathbb{C}/\Lambda$ is a conformal equivalence. On the nonzero half-lattice points it acts bybecause ${\omega }^2=-1-\omega$. Thus it acts transitively.

For a compact subset $C\subset \mathfrak{h}$, the real-linear maphas inverse norm bounded uniformly for $\tau \in C$. Hence there is $A_C>0$ such thatTherefore the absolute value of the defining series is bounded locally uniformly bywhich converges for $k>2$. The Weierstrass M-test gives locally uniform absolute convergence, so termwise holomorphy proves that $G_k^{(x,y)}$ is holomorphic.