# Best polynomial approximation of a partially constant function

Suppose $$\alpha \in [0;1]$$. Let's define $$f_\alpha$$ as the function $$[0;1] \to [0;1]$$ with the following formula:

$$f_\alpha(x) = \begin{cases} 0 & \quad x < \alpha \\ 1 & \quad x \geq \alpha \end{cases}$$

What is the best $$n$$-degree polynomial approximation of $$f_\alpha(x)$$?

By best $$n$$-degree polynomial approximation I mean a polynomial $$p_n(x)$$ of degree $$\leq n$$ such that $$\sup_{x \in [0;1]} |f_\alpha(x) - p_n(x)|$$ is smallest possible.

The best polynomial approximation exists due to the facts that $$L_\infty[0;1]$$ is a Banach space, all polynomials of degree at most $$n$$ form its closed convex subspace and there is a projection theorem for Banach spaces.

## 1 Answer

For all $$n$$, the constant polynomial $$c(x) = \frac{1}{2}$$ is the best approximation for the function $$f_{\alpha}$$.

To see this, we show that for every polynomial $$p : [0,1] \to \Bbb R$$ the following holds: $$\sup_{x \in [0,1]} |f_{\alpha} (x) - c(x)| \le \sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|$$

First: obviously $$\sup_{x \in [0,1]} |f_{\alpha} (x) - c(x)|= \frac{1}{2}$$

So it is enough to prove that for any polynomial $$p$$ the inequality $$\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|\ge \frac{1}{2}$$ holds. There are two cases:

1. First case: $$p( \alpha ) \le 1/2$$. In this case you have $$|f_{\alpha} (\alpha) - p(\alpha)| = |1-p(\alpha)| = \left|\frac{1}{2} + \frac{1}{2} - p( \alpha) \right| = \frac{1}{2} + \left( \frac{1}{2} - p( \alpha) \right) \ge \frac{1}{2}$$

In particular the supremum is larger or equal to $$1/2$$.

1. Second case: $$p( \alpha ) > 1/2$$. In this case consider arbitrary $$x \in [0 ;\alpha )$$. You have $$\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)| \ge \sup_{x \in [0,\alpha)} |f_{\alpha} (x) - p(x)| \ge \lim_{x \to \alpha^-} |f_{\alpha} (x) - p(x)| = p(\alpha) > \frac{1}{2}$$

This completes the proof.

Remark: the polynomial minimizing the quantity $$\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|$$ need not to be unique. There may be more than one, but I think that the constant polynomial is the simplest one.