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Let $f:\mathbb R \rightarrow\mathbb R $ be continuous, and choose $-\infty < a<b<\infty$ and $\epsilon > 0$. Show there exists a polynomial $p$ such that:

a) $p(a)=f(a),\; p(b)=f(b)$

b) $|f(t)-p(t)|<\epsilon$ for all $t\in [a,b]$

The second part is just Weierstrass approximation theorem. But the first part I didn't manage to get. I tried conjuring up a polynomial in the form $$p=p_\epsilon+(\frac{a-x}{a-b})^Nf(b)+(\frac{b-x}{b-a})^Nf(a)$$ such that $p_\epsilon$ satisfies b by the theorem, and then play with constants to set bounds on the maximal error that the extra terms introduce (and also to cancel out $p_\epsilon$ at the bounds). I don't think this can work, since it eventually means that the extra terms are expected to approximate "some portion" of $f$, but given the fact that the proof to Weiersttrass' theorem is pretty complicated, I guess this isn't the way to go... How should I approach this properly? Thanks

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  • $\begingroup$ Note that we can reduce the problem to the case $[a,b]=[0,1]$ since we can define $g:[0,1]\to[a,b]$ by $g(t)=(1-t)a+tb$ then find a polynomial $p$ that satisfies $(1)$ and $(2)$ on $[0,1]$. Then the polynomial $p\circ g^{-1}$ gives the desired conclusion. $\endgroup$ – user71352 Jan 3 '17 at 0:36
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    $\begingroup$ For the case of $[0,1]$ note that there is a proof of Weierstrass' Theorem using Bernstein Polynomials in which the approximating polynomial is $B_{n}(f)(x)=\sum_{\nu=0}^{n}f\left(\frac{\nu}{n}\right)b_{\nu,n}(x)$ where $b_{\nu,n}$ are polynomials that have the property $b_{\nu,n}(0)=\delta_{\nu,0}$ and $b_{\nu,n}(1)=\delta_{\nu,n}$. I think this gives the desired conclusion. $\endgroup$ – user71352 Jan 3 '17 at 0:36
  • $\begingroup$ To reiterate, Bernstein Polyomials can be use to constructively prove the theorem (so, if you follow along with your specific case, you'll end up with the correct answer written down), with your additional attributes that $p(0) = p(1) = 1$. $\endgroup$ – Mark Jan 3 '17 at 0:39
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Let's do it first for the case $f(a) = f(b)=0.$ Find polynomials $P_n \to f$ uniformly on $[a,b].$ Then verfify that the polynomials

$$Q_n(x)=P_n(x)- P_n(a) -\frac{P_n(b)-P_n(a)}{b-a}(x-a)$$

converge to $f$ uniformly on $[a,b],$ with $Q_n(a) = Q_n(b) = 0$ for all $n.$

In the general case, let $l(x)$ be the linear function connecting $(a,f(a))$ to $(b,f(b)).$ Then $f(x) - l(x)$ equals $0$ at the endpoints. From the above we can find polynomials $Q_n \to f-l$ uniformly on $[a,b],$ with $Q_n(a)= Q_n(b) = 0.$ Now check that $Q_n + l$ satisfies the requirements.

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You can prescribe equality of the functions at an arbitrary finite set of points ( derivatives too if you like). Let $P$ approximating $f$ by at most $\delta$. Your new polynomial should be $$P + Q$$ where $Q$ interpolates $f-P$ at the given points. The values of $f-P$ are all $\le \delta$. We can choose $\delta$ so that $Q$ does not exceed $\epsilon/2$ on the whole set. Make sure to take $\delta \le \epsilon/2$ also. Then $P+Q$ does not differ from $f$ by nore than $\epsilon$ and coincides with $f$ at the given points.

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