I have $p$ a polynomial given by $p(x) = a_0 + a_1 x + a_2 x^2 ... a_n x^n$ and a linear operator $T$ defined by $T(p)(x) = a_0 + a_1 x^2 + a_2 x^4 + ... + x^{2n}$. The norm on the space is given by $||p|| = \sup \{|p(t)| :t\in[0,1] \}$.
I'm trying to determine if $T$ is bounded but I'm not having much luck and I'm actually not sure what the correct answer should be.
It's clear that we have $x^2 \le x$ on $[0,1]$ and so on the "basis" $\{x^n\}$ $T$ is bounded but I'm aware this isn't good enough to show that T is bounded.
I've also tried to construct counter examples, i.e. sequences of polynomials $p_n$ where $\frac{||T(p_n)||}{||p_n||}$ can be made arbitrary large. My thoughts here were to try things that become unbounded when $x \to x^2$ e.g. $p_n = n - nx$ but this is not working for me either.
Thanks for any help
Doesn't this operator has norm 1 ?
$$\| Tp \| = \sup \{ |p(t^2)| : t \in [0,1] \} = \sup \{ |p(x)| : x \in [0,1] \} = \|p\|$$