# Best approximation by polynomials.

Let $P_n$ be a polynomial of degree $n$. For a bounded function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$, where the infimum is taken all polynomials of degree $n$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.

I'd like to prove follwing statement:

If $Q_\lambda(x)$ is the form of $\lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$.

I would prove this by contradiction. So assume that there is no such $\Delta(Q_{\lambda_0})$, then for every $\Delta(Q_{\lambda})$, there is $\Delta(Q_{\lambda_i})$ such that $\Delta(Q_{\lambda_i})<\Delta(Q_{\lambda})$.

And I'm stuck on this point. I can't prove this without assumption that is f is continuous function. What step is needed to prove? Or Is there a constructive argument to prove? Any advice would be welcomed!

1. prove that $$\Delta(\lambda)=\sup_{x\in[a,b]}|f(x)-\lambda P_n(x)|$$ is a function that is uniformly continuous on $$\mathbb{R}$$ (and so continuous in every real number);
2. show that $$\Delta(\lambda)$$ is a strictly growing function as $$x\to+\infty$$ and as $$x\to-\infty$$;
3. use Weierstrass theorem for continuous functions in closed intervals to conclude.

Here is my attempt to prove step 1 from @Nameless's answer:

1. Let $$f(x), g(x), h(x)$$ be arbitrary bounded functions on $$[a,b]$$. We can notice that :

$$\sup _{x\in[a,b]} |f(x)-g(x)| \le \sup _{x\in[a,b]} |f(x)-h(x)| + \sup _{x\in[a,b]} |g(x)-h(x)| \tag{1}\label{eq1}$$

Proof of $$\ref{eq1}$$:

Now we can use triangle inequality to see that $$\forall c \in [a,b]$$: \begin{aligned} |f(c)-g(c)| &\le |f(c)-h(c)| + |g(c)-h(c)| \\ &\le \sup _{x\in[a,b]} |f(x)-h(x)| + \sup _{x\in[a,b]} |h(x)-g(x)| \end{aligned} \tag{2}

So $$\sup _{x\in[a,b]} |f(x)-h(x)| + \sup _{x\in[a,b]} |h(x)-g(x)|$$ is upper bound of subset of real numbers $$\{ |f(x)-g(x)| \mid x \in [a,b]\}$$ and cannot be less than $$\sup _{x\in[a,b]} |f(x)-g(x)|$$

That is why holds $$\ref{eq1}$$ (It is also called triangle inequality for supremum metric)

1. Now let $$\lambda_0$$ and $$\lambda_1$$ be arbitrary real numbers. Denote $$\lambda_0 P_n(x)$$ as $$g$$, $$\lambda_1 P_n(x)$$ as $$h$$. From $$\ref{eq1}$$ follows:

$$\sup _{x\in[a,b]} |f(x)-\lambda_0 P_n(x)| \le \sup _{x\in[a,b]} |f(x)-\lambda_1 P_n(x)| + \sup _{x\in[a,b]} |\lambda_0 P_n(x)-\lambda_1 P_n(x)| \tag{3}\label{eq3}$$

$$\sup _{x\in[a,b]} |f(x)-\lambda_0 P_n(x)| - \sup _{x\in[a,b]} |f(x)-\lambda_1 P_n(x)| \le \sup _{x\in[a,b]} |\lambda_0 P_n(x)-\lambda_1 P_n(x)| \tag{4}\label{eq4}$$

Now without loss of generality we can assume that $$\lambda_0$$ and $$\lambda_1$$ are chosen in such way that: $$\sup _{x\in[a,b]} |f(x)-\lambda_0 P_n(x)| \ge \sup _{x\in[a,b]} |f(x)-\lambda_1 P_n(x)| \tag{5}\label{eq5}$$

Now left part of \ref{eq4} can be rewritten as (in notation of @Nameless):

$$|\sup _{x\in[a,b]} |f(x)-\lambda_0 P_n(x)| - \sup _{x\in[a,b]} |f(x)-\lambda_1 P_n(x)| | = |\Delta(\lambda_0) - \Delta(\lambda_1)| \le \\ \le \sup _{x\in[a,b]} |\lambda_0 P_n(x)-\lambda_1 P_n(x)| = |(\lambda_0 - \lambda_1) \sup_{x\in[a,b]}P_n(x)| \tag{6}\label{eq6}$$

Last supremum exists because of continuity of polynomials and compactness of $$[a,b]$$. Denote it as $$s$$. Now $$|\Delta(\lambda_0) - \Delta(\lambda_1)| \le s( \lambda_0-\lambda_1) \tag{7}\label{eq7}$$ \ref{eq7} $$\Rightarrow \Delta(\lambda)$$ is uniformly continuous on $$\mathbb{R}$$

q.e.d