Affine transformation inside of a function on $C[a,b]$.

Question

Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$.

Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$

Prove that $T_\sigma$ satisfies the following:

a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$

b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$

c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$

d) $||T_\sigma(f)||=||f||$

e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$

This is a problem from "A Short Course on Approximation Theory." It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$

Answer 1

For example, to prove a), take any $t\in[0, 1]$; then

$$(T_\sigma(f+g))(t)=(f+g)(\sigma(t))=f(\sigma(t))+g(\sigma(t))=(T_\sigma(f))(t)+(T_\sigma(g))(t)=(T_\sigma(f)+T_\sigma(g))(t)$$ so $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$.

Answer 2

The first one is easy: $$(T_{\sigma}(f+g))(t)=(f+g)(\sigma(t))=f(\sigma(t))+g(\sigma(t))=(T_{\sigma}(f))(t)+(T_{\sigma}(g))(t)=(T_{\sigma}(f)+T_{\sigma}(g))(t)$$ From that we can conclude $T_{\sigma}(f+g)=T_{\sigma}(f)+T_{\sigma}(g)$.

b) and c) is similar to a) and also easy!

d) $\lVert T_{\sigma}(f)\rVert=\max_{0\leq t\leq 1}|(T_{\sigma}(f))(t)|=\max_{0\leq t\leq 1}|f(\sigma(t)|=\lVert f\rVert$

e) is for the reader