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For a strictly decreasing positive sequence ${ϵ}_n\to 0$, put $a_0={ϵ}_0-{ϵ}_1$ and $a_k={ϵ}_{3^{k-1}}-{ϵ}_{3^k}$ for $k\ge 1$. These are positive with sum ${ϵ}_0$. The positive lacunary Chebyshev series $f_{ϵ}={\sum }_k{a}_k{T}_{3^k}$ has error ${ϵ}_0$ at degree zero and error ${ϵ}_{3^{K-1}}\ge {ϵ}_n$ whenever $3^{K-1}\le n<3^K$. This proves the lower-bound form of Bernstein's lethargy theorem with an explicit continuous function.

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.

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.