Let $a>1$ and $\mathbb{P}^a_n = \{p \in \mathbb{P}_n: p(a)=1 \}$. Define $\hat{p}_n \in \mathbb{P}^a_n$ by $ \hat{p}_n = \frac{T_n(x)}{T_n(a)}$, where $T_n$ is the Chebyshev polynomial of degree $n$ and let $\|\cdot \|_\infty $ denote the max norm on the interval $[1,-1]$. Prove: $$ \|\hat{p}_n\|_\infty \leq \|p\|_\infty,\quad \text{ for all }\quad p \in \mathbb{P}^a_n.$$

I'm having a terrible time figuring out how to go about doing this. I've been trying to do a proof by contradiction by assuming that $\|\hat{p}_n\|_\infty > \|p\|_\infty$. I've rewritten

$$p= \frac{(x-t_1)(x-t_2) \dotsm (x-t_n)}{(a-t_1)(a-t_2) \dotsm (a-t_n)},$$

where the $t_i$'s are zeros of the polynomial and have ended up with this inequality:

$$ \frac{1}{|T_n(a)|} \max_{x \in [-1,1]} |T_n(x)| > \frac{1}{\left|(a-t_1)(a-t_2) \dotsm (a-t_n)\right|} \max_{x \in [-1,1]} \left|(x-t_1) \dotsm (x-t_n)\right|.$$ I'm unsure of what to do next. I feel as though I should be able to make some contradiction concerning the number of zeros, but I'm not seeing it. Any type of insight would be incredibly helpful. Thank you!


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