Is there any example demonstrating nonlinearity of best polynomial approximation operator?
For any $fin C[0,1]$, it is well known that there exists an unique $p^{*}in P_n[0,1]$ such that $||f-p^{*}||_{infty}=inflimits_{pin P_n[0,1]}||f-p||_{infty}$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as $A_n(f)=p^{*}$. Can any one provide an example such that $f_1,f_2in C[0,1]$ and $A_n(f_1+f_2)neq A_n(f_1)+A_n(f_2)?$
polynomials approximation-theory
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For any $fin C[0,1]$, it is well known that there exists an unique $p^{*}in P_n[0,1]$ such that $||f-p^{*}||_{infty}=inflimits_{pin P_n[0,1]}||f-p||_{infty}$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as $A_n(f)=p^{*}$. Can any one provide an example such that $f_1,f_2in C[0,1]$ and $A_n(f_1+f_2)neq A_n(f_1)+A_n(f_2)?$
polynomials approximation-theory
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For any $fin C[0,1]$, it is well known that there exists an unique $p^{*}in P_n[0,1]$ such that $||f-p^{*}||_{infty}=inflimits_{pin P_n[0,1]}||f-p||_{infty}$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as $A_n(f)=p^{*}$. Can any one provide an example such that $f_1,f_2in C[0,1]$ and $A_n(f_1+f_2)neq A_n(f_1)+A_n(f_2)?$
polynomials approximation-theory
For any $fin C[0,1]$, it is well known that there exists an unique $p^{*}in P_n[0,1]$ such that $||f-p^{*}||_{infty}=inflimits_{pin P_n[0,1]}||f-p||_{infty}$. In this fashion, one can define an operator $A_n: C[0,1]mapsto P_n[0,1]$ as $A_n(f)=p^{*}$. Can any one provide an example such that $f_1,f_2in C[0,1]$ and $A_n(f_1+f_2)neq A_n(f_1)+A_n(f_2)?$
polynomials approximation-theory
polynomials approximation-theory
asked Nov 20 '18 at 5:53


Lin Xuelei
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