Suppose that $f \in C[-1,1]$ has a chebyshev series $\sum_{n=1}^{\infty}a_nT_n$
(b) show that $E_n(T_{n+1})=1$
(c) show that $|E_n(f)-|a_{n+1}|| \le \sum_{k=n+2}^{\infty}|a_k|$
cf : $E_n(f)= \inf\{||f-q||_{\infty} : q \in P_n\}$ where $P_n$ is vector space that contain all the polynomial degree at most $n$
I already solved the question above (b). But I could solve above (c).
my attempt : It already show $E_n(f)-|a_{n+1}| \le \sum_{k=n+2}^{\infty}|a_k|$
So we are only to prove opposite side of inequality
Notice by above (b) we can induce $E_n(|a_{n+1}|T_{n+1})=|a_{n+1}|$
thus If we can show $E_n(f)\ge E_n(|a_{n+1}|T_{n+1})-\sum_{k=n+2}^{\infty}|a_k|$
we can get results
But I cannot show how to solve subsequent process. Please give me a hint ! Thank you.
The key is the obvious inequality $$E_n(f+g)\le E_n(f)+E_n(g).$$ Replacing $f$ by $f-g$, we get $$E_n(f)\le E_n(f-g)+E_n(g),$$ i.e. $$E_n(f)-E_n(g)\le E_n(f-g).$$ Exchanging $f$ and $g$ gives $$E_n(g)-E_n(f)\le E_n(f-g),$$ and both together $$|E_n(f)-E_n(g)|\le E_n(f-g),\tag1$$ since $|x|=\max(x,-x)$. Moreover, we have $E_n(g)=E_n(g+p)$ for any polynomial $p\in P_n$ by definition. Now set $$g=\sum^{n+1}_{k=0}a_k\,T_k=\sum^n_{k=0}a_k\,T_k+a_{n+1}T_{n+1}.$$ Then, $$E_n(g)=E_n(a_{n+1}T_{n+1})=|a_{n+1}|,\tag2$$ and $$E_n(f-g)=E_n\left(\sum^\infty_{k=n+2}a_k\,T_k\right)\le\sum^\infty_{k=n+2}|a_k|,\tag3$$ and you just have to plug (2) and (3) into (1).