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!


Follow these steps:

  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 $\mathbb{R}$



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