Suppose $f$ is a continuous fonction on an interval $[a,b]$. Given $\varepsilon > 0$ and $n \in \mathbb{N}$, does there exists a polynomial $p$ on $[a,b]$ such that $|f(x) - p(x)| < \epsilon$ for all $x$ in $[a,b]$ and $|p'(x)| < \epsilon$ for all $x$ where $|f(x)|< \frac{1}{n}$ ?
Approximation by a polynomial with small derivative on a neighbourhood of zero
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real-analysis
analysis
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2What does it mean that $|f|$ is small? What is the quantifier for $\varepsilon$? – 2017-02-16
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0I hope my question is clearer now. – 2017-02-16
1 Answers
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Not necessarily. Suppose $(2+b-a) \epsilon < 1/n$. Then you can take
$$f(x) = c \frac{x-a}{b-a}$$ where $$ (2+b-a) \epsilon < c < 1/n $$
Suppose polynomial $p$ satisfies your conditions. Since $|f(x)| \le c < 1/n$
everywhere, $|p'(x)| < \epsilon$ everywhere, and $$|p(b) - p(a)| \le \int_a^b p'(x)\; dx < (b-a)\epsilon$$ But we need $$|p(b) - p(a)| > f(b) - f(a) - 2 \epsilon = c - 2 \epsilon > (b-a)\epsilon$$ contradiction!