How to prove the best polynomial approximation operator is continuous?
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For any function $fin C[0,1]$, it is well known that there exist an unique polynomial $p^{*}in P_n[0,1]$ such that $||p^{*}-f||_{infty}leq ||p-f||_{infty}$ for any $pin P_n[0,1]$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as
$$
A_n(f):={rm argmin}_{pin P_n[0,1]}||f-p||_{infty}.
$$
For a fixed $fin C[0,1]$, how to prove that there exists a constant $C$ depending on $f$ and $n$ such that
$$
||A_n(f)-A_n(tilde{f})||_{infty}leq C||f-tilde{f}||_{infty},quad foralltilde{f}in C[0,1].
$$
functional-analysis approximation-theory nonlinear-analysis
asked Nov 18 at 15:45
Lin Xuelei
10810
add a comment |
up vote
1
down vote
favorite
For any function $fin C[0,1]$, it is well known that there exist an unique polynomial $p^{*}in P_n[0,1]$ such that $||p^{*}-f||_{infty}leq ||p-f||_{infty}$ for any $pin P_n[0,1]$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as
$$
A_n(f):={rm argmin}_{pin P_n[0,1]}||f-p||_{infty}.
$$
For a fixed $fin C[0,1]$, how to prove that there exists a constant $C$ depending on $f$ and $n$ such that
$$
||A_n(f)-A_n(tilde{f})||_{infty}leq C||f-tilde{f}||_{infty},quad foralltilde{f}in C[0,1].
$$
functional-analysis approximation-theory nonlinear-analysis
asked Nov 18 at 15:45
Lin Xuelei
10810
Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For any function $fin C[0,1]$, it is well known that there exist an unique polynomial $p^{*}in P_n[0,1]$ such that $||p^{*}-f||_{infty}leq ||p-f||_{infty}$ for any $pin P_n[0,1]$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as
$$
A_n(f):={rm argmin}_{pin P_n[0,1]}||f-p||_{infty}.
$$
For a fixed $fin C[0,1]$, how to prove that there exists a constant $C$ depending on $f$ and $n$ such that
$$
||A_n(f)-A_n(tilde{f})||_{infty}leq C||f-tilde{f}||_{infty},quad foralltilde{f}in C[0,1].
$$
functional-analysis approximation-theory nonlinear-analysis
asked Nov 18 at 15:45
Lin Xuelei
10810
For any function $fin C[0,1]$, it is well known that there exist an unique polynomial $p^{*}in P_n[0,1]$ such that $||p^{*}-f||_{infty}leq ||p-f||_{infty}$ for any $pin P_n[0,1]$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as
$$
A_n(f):={rm argmin}_{pin P_n[0,1]}||f-p||_{infty}.
$$
For a fixed $fin C[0,1]$, how to prove that there exists a constant $C$ depending on $f$ and $n$ such that
$$
||A_n(f)-A_n(tilde{f})||_{infty}leq C||f-tilde{f}||_{infty},quad foralltilde{f}in C[0,1].
$$
functional-analysis approximation-theory nonlinear-analysis
functional-analysis approximation-theory nonlinear-analysis
asked Nov 18 at 15:45
Lin Xuelei
10810
asked Nov 18 at 15:45
Lin Xuelei
10810
asked Nov 18 at 15:45
Lin Xuelei
10810
asked Nov 18 at 15:45
Lin Xuelei
10810
asked Nov 18 at 15:45
Lin Xuelei
10810
10810
Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20
add a comment |
Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20
Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20
add a comment |
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Is $A_n$ linear?
– Paul Frost
Nov 18 at 16:18
No, it is not linear, since the underlying norm is supremum norm.
– Lin Xuelei
Nov 18 at 17:20