Chebyshev coefficients upper bounded by sup-norm of function?

Question

Let $T_k(x)$ be the Chebyshev polynomials of the first kind and consider the function $$ f(x) = \sum_{k = 0}^\infty c_k \, T_k(x). $$ Show that $$ |c_k| \leq \|f\|_{[-1,1]}. $$

Answer 1

By the definition of Chebyshev coefficients, we have that $$ \begin{aligned} |a_k| &= \left| \frac{2}{\pi} \int_{-1}^1 \frac{f(x) \, T_k(x)}{\sqrt{1-x^2}} \, dx \right| \\&\leq \|f(x) \, T_k(x)\|_{[-1,1]} \, \left| \frac{2}{\pi} \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} \, dx \right| \\&\leq \|f(x)\|_{[-1,1]} \end{aligned} $$ where on the last line I used that $\|T_k(x)\|_{[-1,1]} = 1$ and $$ \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} \, dx $$ is the area $\frac{\pi}{2}$ of half the unit disk.