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First I'd like to say that although this question was asked before (here) and is from the same text, the answer used methods that were not introduced in the text.

Let $P_n(x)$ be a polynomial of degree $n$ and $f:[a,b] \to \Bbb R$ a bounded function.

Define $$\Delta(P_n) = \sup_{x\in[a,b]}|f(x) -P_n(x)| \ \ \mathrm{and} \ \ E_n(f) =\inf_{P_n} \Delta(P_n),$$ the latter being taken over all polynomials of degree $n$. A polynomial is called a polynomial of best approximation of $f$ if $\Delta(P_n) = E_n(f)$

The previous questions are:

$a)$ show a polynomial of best approximation of degree $0$ exists;

$b)$ among the polynomials $Q_{\lambda}(x)$ of the form $\lambda P_n(x)$, where $P_n$ is a fixed polynomial, there is a polynomial $Q_{\lambda_0}$ such that $$\Delta(Q_{\lambda_0}) = \min_{\lambda \in \Bbb R}\Delta(Q_{\lambda}).$$

The question I am on is:

$c)$ if there exists a polynomial of approximation of degree $n$, there also exists a polynomial of best approximation of degree $n + 1$.

This last part looks like an inductive step which would use $b)$ with $f$ as $f(x) - P_n(x)$ and $Q_{\lambda}$ as $\lambda x^{n+1},$ $P_n(x)$ being the best approximation. In general, though, the polynomial of degree $n+1$ is not in the form of $P_n(x) + \lambda_0 x^{n+1}.$ Because of this I'm not sure how the question can be done.

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  • I'm not even sure if the claim is correct. It seems to me that there only exists a best approximation of degree $\le n+1$. I.e. if $f$ is a $1$st degree polynomial, the best approximation of degree $2$ should have the coefficient of $x^2$ equal to $0$, in which case it wouldn't really be of degree $2$. – J. J. Apr 23 '14 at 11:11
  • @J.J. I also thought of that counter example so we know induction doesn't work. But as I said, the question is set up to use induction so I don't know how to proceed. – user110503 Apr 23 '14 at 11:15
  • @J.J. With that example, there is a polynomial of best approximation of any degree ( I think), but it can't be $\lambda x^2 + ax$ because of what you said. – user110503 Apr 23 '14 at 11:18
  • @user110503: To be precise, say that $f(x) = x$ on $[0,1]$. Then taking $\frac{1}{m} x^2 + x$ with $m \to \infty$ you find that $E_2(f) = 0$ so that if a best approximation did exist, it would equal $f$ on the entire interval, and in particular it's not of degree $2$. – fuglede Apr 23 '14 at 11:21

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