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We need to show that for all $f: [a,b] \rightarrow \Bbb R $ which is differentiable $n$ times, and for all $\varepsilon>0$ there exists a polynomial $p\colon[a,b] \rightarrow R$ s.t. $\forall n \geqslant k\geqslant 0, \forall x\in [a,b] : |f^{(k)}(x)-p^{(k)}(x)|<\varepsilon$

I thought I'd take $p$ as the Taylor series of $f$ and show that it's a polynomial and equals $f$ when the remainder of it converges to zero and then because of equality the derivatives are equal? But I'm really not sure about it... Any leads? :]

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  • $\begingroup$ You mean, you seek a single polynomial for all $0 \leq k \leq n$, or a polynomial $p_k$ for each $k$? $\endgroup$
    – xyzzyz
    Jun 9, 2013 at 15:28
  • $\begingroup$ Its a single polynomial for every $\epsilon$ so basically it would be $p_\epsilon$ $\endgroup$ Jun 9, 2013 at 15:29
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    $\begingroup$ Your statement at least implies that $f^{(n)}$ is bounded, but this could fail in general. Do you assume that $f^{(n)}$ is continuous? $\endgroup$
    – 23rd
    Jun 9, 2013 at 15:44
  • $\begingroup$ doesn't say anything about continuity.. $\endgroup$ Jun 9, 2013 at 16:40

2 Answers 2

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We may assume $[a,b]=[-1,1]$; furthermore we assume that $f\in C^n\bigl([-1,1]\bigr)$, where $n\geq0$. The following is a proof by induction.

When $n=0$ (i.e., $f$ is continuous on $[-1,1]$) and an $\epsilon>0$ is given then according to the Stone-Weierstrass theorem we can find a polynomial $p$ such that $$|f(x)-p(x)|\leq\epsilon\qquad(-1\leq x\leq 1)\ .$$ (Such a polynomial has nothing to do with any Taylor expansion the function $f$ might have.)

Assume now that the statement is true for $n-1\geq0$ and that $f\in C^n\bigl([-1,1]\bigr)$. Applying it to $f'$ we obtain a polynomial $p$ whose derivatives up to order $n-1$ approximate the corresponding derivatives of $f'$: $$|(f')^{(k)}(x)- p^{(k)}(x)|<\epsilon \qquad(0\leq k\leq n-1)\ .\tag{1}$$ We now define the polynomial $P$ by $$P(x):=f(0)+\int_0^x p(t)\ dt\qquad(-1\leq x\leq 1)\ .$$ Then $$|f(x)-P(x)|\leq\left|\int_0^x \bigl(f'(t)-p(t)\bigr)\ dt\right|\leq |x|\sup_{-1\leq t\leq1}|f'(t)-p(t)|<\epsilon\ ,$$ and using $(1)$ it is easy to see that $P$ satisfies stated requirements for $f$.

Without the assumption that $f^{(n)}$ is continuous the stated claim is false. A sequence of polynomials $(q_k)_{k\geq1}$ converging uniformly to some $g$ on the interval $[-1,1]$ necessitates that $g$ is continuous to begin with. As an example consider the function $$f(x):=\cases{x^2\sin{1\over x}\quad&$(0<|x|\leq 1)$ \cr 0&$(x=0)$ .\cr}$$ This function is differentiable on all of $[-1,1]$, but as $$f'(x)=2x\sin{1\over x}-\cos{1\over x}\qquad(x\ne0)$$ there is no polynomial that can approximate $f'$ with an error $<1$ in the $\sup$-norm.

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  • $\begingroup$ It's not that easy to see, am I to understand that its a proof by induction? if so how does choosing the $p$ given by Weierstrass on $f'$ and creating $P(x)$ that way satisfies the induction and the requirement?.. $\endgroup$ Jun 10, 2013 at 10:45
  • $\begingroup$ @AddarBokobza: See my Edit. $\endgroup$ Jun 10, 2013 at 11:16
  • $\begingroup$ How to argue that $f'(x)$ in your last paragraph cannot be approximated by an polynomial with an error $<1$? All I can see is that $f'(x)$ is not continuous at $0$ thus cannot be the uniform limit of a sequence of polynomials on $[-1,1]$. $\endgroup$
    – Bach
    Aug 12, 2019 at 12:44
  • $\begingroup$ @ChristianBlatter Then how to prove it? I think this is stronger than claiming that there is not a uniform limit of a sequence of polynomials on $[-1,1]$ converging to $f'$. $\endgroup$
    – Bach
    Aug 12, 2019 at 18:27
  • $\begingroup$ @Bach: We want a single polynomial $p$ (not a sequence of polynomials) that approximates $f'$ with an absolute error $≤{1\over2}$ on $[{-1},1]$. Any such $p$ would vertically move less than ${1\over4}$ in a suitable interval $[{-\delta},\delta]$, hence cannot approximate $f'$ in the desired way. $\endgroup$ Aug 12, 2019 at 18:36
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I will assume that $f\in C^\infty[a,b]$, other case can be fixed in a similar way.

Using Stone-Weierstrass theorem and the fact that $f^{(d)}$ is continuous for each $d$, can see that for each $k$, there is a polynomial $P_k$ such that $$\max_{0\leqslant d\leqslant k}\max_{x\in [a,b]}|f^{(d)}(x)-P^{(d)}_k(x)|\leqslant \frac 1k.$$ The key is that (when $k=1$), we take $\lVert f'-P\rVert_\infty$ small, then we consider $\int_a^xP(t)dt$, a polynomial.

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