Question on the norms

Question

I got stuck with the following (simple) question since the result I got seems to be counterintuitive:

I have a function defined in terms of its Chebyshev expansion, i.e.

$\psi(x)=\sum_{i}a_{i}T_{i}(x)$

with $T_{n}(x)$ Chebyshev polynomial of order $n$. Then I have a norm given by $||\psi||=\sum_{i}|a_{i}|\cdot w^{i}$, with $w>1$, weight.

Now, if I rescale my argument by a number $\lambda\in\mathbb{R}$, i.e. consider

$\psi(\lambda x)=\sum_{i}a_{i}T_{i}(\lambda x)$

Do I get then $||\psi(x)||=||\psi(\lambda x)||$ in the norm defined above? Since the coefficients of the actual expansion are the same… I don't know if it makes sense that the norms are equal, would be grateful if anyone could help.

This is not a homework question.

Answer 1

Your argumentation cannot be right, because $T_i(\lambda x) \neq T_i(x)$ in general. For example you have $$T_1(\lambda x)=\lambda x = \lambda T_1(x)$$ and thus $$\|T_1(\lambda x)\|=|\lambda|\cdot\|T_1(x)\|$$ In general you have $$\phi(\lambda x)=\sum b_i T_i(x)$$ with $b_i\neq a_i$ ($a_i$ are the coefficients of $\phi(x)=\sum a_i T_i(x)$).