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Here is a problem that comes from my graduate school entrance exam. Unfortunately, I didn't solve it at the time, and even now, I still have no idea.

Let $f:[2,7]\rightarrow\mathbb{R}$ be continuous. Given $\varepsilon>0$, show that there is a polynomial $P$ such that $$ P(2)=f(2),\quad P'(2)=0,\quad\text{and}\quad\sup\{|P(x)-f(x)|:x\in[2,7]\}<\varepsilon. $$

Due to the way it states, I immediately thought of the so-called Weierstrass approximation theorem:

There is a polynomial $P$ such that $||P-f||_\infty<\varepsilon$,

where $||\cdot||_\infty$ denotes the sup norm. I'm eager to work this out. Any suggestion is welcomed. Thanks.

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For sake of convenience we may replace $[2,7]$ by $[0,1]$. $f(\sqrt x)$ is a continuous function on $[0,1]$, hence by Weierstrass theorem we may find a polynomial $Q(x)$ such that $$\sup_{x\in[0,1]}|f(\sqrt x)-Q(x)|<\epsilon/2$$ Let $P(x)=Q(x^2)+f(0)-Q(0)$. Then $P(x)$ is a polynomial satisfying $P(0)=f(0)$, $P'(0)=0$, and \begin{align} \sup_{x\in[0,1]}|f(x)-P(x)|&\le\sup_{x\in[0,1]}|f(x)-Q(x^2)|+|f(0)-Q(0)|\\ &=\sup_{\sqrt{x}\in[0,1]}|f(\sqrt x)-Q(x)|+|f(0)-Q(0)|\\ &=\sup_{x\in[0,1]}|f(\sqrt x)-Q(x)|+|f(0)-Q(0)|\\ &\le2\sup_{x\in[0,1]}|f(\sqrt x)-Q(x)|<\varepsilon \end{align}

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By Weierstrass theorem, we may find a polynomial $Q$ such that $$\sup_{x\in[2,7]}|f(x)-Q(x)|<\epsilon/2.$$

If $Q^\prime(2)= 0$, we’re done. Otherwise, denote $Q^\prime(2)=\alpha$ and consider the polynomials $P_{\alpha,n}(x)= \alpha x (1-x)^n$ defined on the interval $[0,1]$.

You’ll verify that $P_{\alpha,n}^\prime(0)= \alpha$ and $$\sup\limits_{x \in [0,1]} \vert P_{\alpha,n}(x) \vert \le \frac{\vert \alpha \vert}{n+1}.$$

Likewise $Q_{\alpha,n}(x) = P_{\alpha,n}(\frac{x-2}{5})$ is defined on $[2,7]$, $Q_{\alpha,n}^\prime(2) = \alpha/5$ and $$\sup\limits_{x \in [2,7]} \vert Q_{\alpha,n}(x) \vert \le \frac{\vert \alpha \vert}{n+1}.$$

Taking $\alpha =5 Q^\prime(2)$ and $n$ such that $\frac{\vert \alpha \vert}{n+1} \le \epsilon/2$,

$$P(x)=Q(x)-Q_{\alpha,n}(x)$$ is a polynomial fulfilling the conditions you were looking for.

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  • $\begingroup$ So basically we just approximate the function with any polynomial and then fix the derivative at the interval endpoint by subtracting another polynomial which is small everywhere. $\endgroup$ – Adayah Jun 10 '18 at 16:05
  • $\begingroup$ @Adayah Yes, this was the way I went towards the solution. Initially, I was looking to substract a Chebyshev polynomials. But it is simpler with the polynomials $P_{\alpha,n}$. $\endgroup$ – mathcounterexamples.net Jun 10 '18 at 16:09

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