Why is the polynomial best approximation to an even function itself even?

Solution 1:

First, we must pin down what we mean by "better approximate." If you are interested in the interval $[-1,1]$ and your given function is $f(x)$, one reasonable definition of approximation error by a second function $g(x)$ is $$\int_{-1}^1 [f(x)-g(x)]^2\,dx.$$

Now suppose $f(x)$ is an even function, and $e(x)$ an even polynomial approximation to $f(x)$. Can we improve the approximation by adding some odd polynomial terms $o(x)$? Let's check:

\begin{align*} \int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx &= \int_{-1}^1 [f(x)-e(x)]^2 - 2[f(x)-e(x)]o(x) + o(x)^2\,dx\\ &= \int_{-1}^1 [f(x)-e(x)]^2\,dx + \int_{-1}^1 o(x)^2\,dx -2\int_{-1}^1[f(x)-e(x)]o(x)\,dx. \end{align*} Now let's use the fact that $f$ and $e$ are even, and $o$ is odd: $$\int_{-1}^1[f(x)-e(x)]o(x)\,dx = -\int_0^1[f(x)-e(x)]o(x)\,dx + \int_0^1 [f(x)-e(x)]o(x)\,dx = 0.$$

Therefore $$\int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx = \int_{-1}^1 [f(x)-e(x)]^2 + \int_{-1}^1 o(x)^2\,dx \geq \int_{-1}^1 [f(x)-e(x)]^2$$ and you were better off without the odd terms.

EDIT: For different norms, you carry out different flavors of the same argument. For instance, for the $\sup$ norm, \begin{align*} \sup_{x\in [-1,1]} |f(x)-e(x)-o(x)| &= \sup_{x\in [0,1]} \max\left(|f(x)-e(x)-o(x)|,|f(x)-e(x)+o(x)|\right)\\ &\geq \sup_{x\in [0,1]} |f(x)-e(x)|\\ &= \sup_{x\in [-1,1]} |f(x)-e(x)|. \end{align*}