# Is there any example demonstrating nonlinearity of best polynomial approximation operator?

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 $$A_n(f_1+f_2)\neq A_n(f_1)+A_n(f_2)?$$

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 $$n=1$$

• 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 $f$. – Orest Bucicovschi Oct 7 at 9:28