1
$\begingroup$

Theorem (Stone-Weierstrass). If $f$ is a continuous complex function on $[a,b]$, there exists a sequence of polynomials $P_n$ such that $$\lim \limits_{n\to \infty }P_n(x)=f(x)$$ uniformly on $[a,b]$. If $f$ is real, then $P_n$ may be taken real.

Rudin wrote that "we may assume, without loss of generality, that $[a,b]=[0,1]$". If we found a sequence of polynomials $\{P_n\}_{n=1}^{\infty}$ which converges uniformly to $f(x)$ on $[0,1]$. How to extend it to $[a,b]$?

Can anyone explain it to me please?

$\endgroup$
2
$\begingroup$

Let $F: [0,1] \to [a,b]$ be a linear bijective map. Then both $F, F^{-1}$ are continuous. For all $f\in C[a,b]$, $f\circ F^{-1}$ is a continuous function on $[0,1]$ and so there is $P_n$ so that $P_n$ converges uniformly to $f\circ F^{-1}$ on $[0,1]$. Then $P_n \circ F$ converges uniformly to $f$ on $[a,b]$. Note that $P_n\circ F$ is also a polynomial as $F$ is linear.

$\endgroup$
  • 1
    $\begingroup$ Now I understand! Thank you very much, dear John! $\endgroup$ – ZFR Nov 28 '15 at 10:03
  • $\begingroup$ Dear John! Can I ask your help in my question? math.stackexchange.com/questions/1551502/… $\endgroup$ – ZFR Nov 29 '15 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.