Consider the expansion of $f \in [-1,1]$ in Chebyshev polynomials of the first kind $T_n(x)$: $$f(x)=\sum_{n=0}^\infty a_{n} T_n(x), \quad a_n=\frac{2}{\pi} \int_{-1}^1 \frac{f(x) T_n(x)}{\sqrt{1-x^2}} \mathrm{~d} x$$ Calculating these coefficients, it seems to me that they decrease with $n$.
If it is possible to calculate this expansion, that is, if the series converges to the function, is it always true that the coefficients are decreasing? Perhaps this is a well known fact, explained in some textbook, but I cannot prove (or disprove) this claim. Could somebody help me please?
Thanks you!
If $\displaystyle \int\limits_{[-1,1]} |f(x)|^2\, \frac{dx}{\sqrt{1-x^2}} < +\infty$ then $\displaystyle\sum_{n\,=\,0}^\infty |a_n|^2<+\infty, $ so we have $a_n\to0$ as $n\to\infty.$
(The absolute-value sign is needed when values of $f(x)$ and hence of $a_n$ are complex numbers that need not be real. If one restricts everything to real numbers, then squares are nonnegative.)