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Let f be a continuously differentiable function on $[a.b]$. Show that there is a sequence of polynomials $\{P_n\}$ such that $P_n(x) \to f$ and $P'_n(x) \to f' (x)$ uniformly on $[a,b]$

My approach has been as follows. Since f is continuously differentiable, we have $Q_n(x) \ to f'(x)$ on$ [a,b]$ uniformly (Weierstrass) .

I'm not sure how to find a polynomial that converges to f uniformly who's derivative is $Q(x)$. My first thought was $P(x)=\int_{a}^{x} Q(t)dt$. There is a relationship between uniform continuity and differentiation, but that requires convergence at a point. I'm having problems building a good argument for this part of the proof.

sarahz
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3 Answers3

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You're on the right track; let $P_n(x) = f(a) + \int_a^x Q_n(y)\,dy$. Then since $P_n(a) = f(a)$ you have $$f(x) - P_n(x) = (f(x) - f(a)) - (P_n(x) - P_n(a))$$ $$=\int_a^x f'(y)\,dy - \int_a^x P_n'(y)\,dy$$ $$= \int_a^x (f'(y) - Q_n(y))\,dy$$ If $|f'(y) - Q_n(y)| < \epsilon$ for all $y$ then the absolute value of this integral is at most $(b-a)\epsilon$, which gives what you want.

Zarrax
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2

A more general result

The Bernstein polynomial is defined on $[0,1]$ by $$B_n(f)(x)=\sum_{i=0}^n f(\frac{i}{n})\binom{n}{i}x^i (1-x)^{n-i}$$ and if $f$ is a function with continuous $k^{th}$ derivative then $$||f^{(k)}-B_n(f)^{(k)}||_\infty\to_{n\to\infty}0$$

Note: we can pass from $[0,1]$ to $[a,b]$ by the affine function $x\mapsto ax+b(1-x)$.

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All the antiderivatives of $Q_n$ are of the form $x\mapsto\int_a^xQ_n+C_n$ for some $C_n\in\Bbb R$. We can choose these $C_n$'s such that $P_n(b)=f(b)$ (or $P_n(a)=f(a)$).

Berci
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