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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.