# Proof check (Abbott Understanding Analysis exercise 6.5.10)

My proof: 6.5.10. We will prove by induction that for each $$k \in \mathbb{N} \cup \{0\}$$, there exists $$(c_n) \to 0$$ such that for each $$n \in \mathbb{N}$$, $$c_n \neq 0$$ and $$g^{(k)}(c_n) = 0$$.

Base case, $$k = 0$$: We are given $$(x_n) \to 0$$, and for each $$n \in \mathbb{N}$$, $$x_n \neq 0$$ and $$g(x_n) = 0$$.

Inductive step: Let $$k \in \mathbb{N} \cup \{0\}$$ be arbitrary and let $$(y_n) \to 0$$ be the sequence corresponding to $$k$$ as in the inductive hypothesis. Since $$g^{(k)}$$ is continuous on $$(-R, R)$$, $$g^{(k)}(0) = \lim_{n \to \infty}g^{(k)}(y_n) = \lim_{n \to \infty}0 = 0$$. We now construct a sequence $$(c_n)$$. For each $$n \in \mathbb{N}$$: By the mean value theorem, there exists $$c \in (0, y_n)$$ such that $$g^{(k + 1)}(c) = \frac{g^{(k)}(y_n) - g^{(k)}(0)}{y_n - 0} = 0$$. Let $$c_n = c$$.

For each $$n \in \mathbb{N}$$, $$0 < c_n < y_n$$. By the squeeze theorem, $$(c_n) \to 0$$. By construction, $$c_n \neq 0$$ and $$g^{(k + 1)}(c_n) = 0$$. This completes the proof by induction.

The previous theorem and the continuity of all derivatives of $$g$$ implies that for each $$n \in \mathbb{N} \cup \{0\}, g^{(n)}(0) = 0$$. This implies that $$g^{(n)}(0) = n!b_n = 0$$ for each $$n \in \mathbb{N}$$, which means $$b_n = 0$$ for each $$n \in \mathbb{N}$$. Thus for each $$x \in (-R, R)$$, $$g(x) = 0$$.

• This is a very nice proof. Mar 22, 2020 at 20:11
• Under the given circumstances one can prove that $g^{(k)} (0)=0$ for all $k=0,1,2,\dots$ (via mean value theorem). This means that $b_n=0$ for all $n$. Mar 23, 2020 at 3:05