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For any $f\in C[0,1]$, it is well known that there exists an unique $p^{*}\in P_n[0,1]$ such that $||f-p^{*}||_{\infty}=\inf\limits_{p\in P_n[0,1]}||f-p||_{\infty}$. 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 $f_1,f_2\in C[0,1]$ and $A_n(f_1+f_2)\neq A_n(f_1)+A_n(f_2)?$

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We'll work instead on $C[-1,1]$, where we have concrete examples.

Consider the Chebyshev polynomials $$T_2(x)=2x^2-1\\ T_3(x)=4x^3 - 3 x$$

So we have $$A_1(2x^2)=1\\ A_1(4x^3)=3x$$ (as polynomial functions).

However, $$A_1(4x^3+2x^2)\ne 3x + 1$$

Indeed, the function $f(x)=4x^3+2x^2 - 3x -1$ has norm $2$, and the maximum value of the modulus is only achieved at $x=1$. So for $\epsilon >0$ small enough we have $\|f-\epsilon \cdot 1\| = 2 - \epsilon < \|f\|$.

This works similarly for any $n$ instead of $n=1$

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  • $\begingroup$ Note that we have a formula for the closest constant function to a function $f$, it is $\frac{\max f + \min f}{2}$, and this is easily seen not be to additive in $f$. $\endgroup$ – Orest Bucicovschi Oct 7 at 9:28

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