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While solving some exercise I came up with this problem:

Assume that $f \in C^2[0,\infty]$ (i.e. $f$ has continuous second derivative on $[0,\infty)$ and there exists limit in $+\infty$ of $f$) be such that f''(0) = 0.

Does there exist a sequence $(f_n) \subset C^2[0,\infty]$ such that $f_n \to f$ and f''_n \to f'' in the supremum norm (i.e. $\lVert f_n - f \rVert_{C[0,\infty]}$, \lVert f''_n - f'' \rVert_{C[0,\infty]} \to 0 ) and all $f_n$'s satisfies f''_n(0) = f'_n(0) = 0?

Maybe I am wrong but I think it would be nearly enough to approximate in that sense linear (in some neighborhood of $0$) functions.

I will be grateful for any hints.

1 Answers 1

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This is not possible. Assume $f(x) = x$ for $ x \le 1$, suitably extended for $x\ge 1$. Now choose $\varepsilon$ small (less than $1/100$, say) and $N_0 \in \mathbb{N}$ such that $|f(x) - f_n(x) | \le \varepsilon, |f^{\prime \prime}(x) - f_n^{\prime \prime}(x)| \le \varepsilon \quad \forall x\in \mathbb{R}, n\ge N_0$.

Your assumption ($f_n^{\prime}(0) = 0$) then implies

$|f_n^{\prime}(x)| = |\int_0^x f_n^{\prime\prime}(t) dt| = |\int_0^x (f_n^{\prime\prime} (t) - f^{\prime\prime}(t) ) dt|\le \varepsilon x \quad \forall x\ge 0 \quad (!)$ since $f^{\prime\prime} = 0$. This allows you to estimate $f_n(0) - f_n(x)$ (integrate once more). Hence $\varepsilon \ge f(x) - f_n(x) = f(x) - f(0) + f(0) - f_n(0) + f_n(0) - f_n(x) \ge x - \varepsilon -(\varepsilon/2) |x|^2 $ Now choose $x = 1/2$, say.

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    Thanks! Really nice. As I thought, linear functions were crucial.2011-12-08