Exercise 6.5.10

Let $f^{(n)}$ denote the $n$’th derivative of $f$ (with $f^{(0)}=f$). The intermediate claims we make along the way are:

1.

If a differentiable function $f$ has a sequence $(x_n)\to 0$ satisyfing $f(x_n)=0$, then its derivative also has a sequence $(y_n)\to 0$ satisfying $f^{\prime }(y_n)=0$.

2.

Any function $f$ with a bounded derivative over an interval containing 0 with some sequence $(x_n)\to 0$ satisfying $f(x_n)=0$, will also satisfy $f(0)=0$.

3.

Given a power series $f(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{a}_n{x}^n$, if $f^{(n)}(0)=0$, then $a_n=0$.

For claim 1, we apply the Mean Value Theorem to get some $y_n$ between $x_n$ and $x_{n+1}$ with $f^{\prime }(y_n)=0$; we thus have $\left|y_n\right|\le \max <!--\; FUNCTION\; APPLICATION\; -->\{x_n,x_{n+1}\}$ and therefore $(y_n)\to 0$.

For claim 2, suppose that $f(0)=𝜖\ne 0$, and $\left|f^{\prime }(x)\right|<M$. Now since $(x_n)\to 0$ we can find some $x_i$ satisfying $\left|x_i\right|<𝜖∕M$. By the Mean Value Theorem, we then have that

for some $c$, violating the assumption that $f^{\prime }(x)$ is bounded. Hence $f(0)=0$.

For claim 3, we differentiate termwise $n$ times and note that all terms that still have $x$ will evaluate to 0. We thus have

and thus $a_n=0$.

From claim 1, we have by indution that every $g^{(i)}$ has some sequence $(x_{i,n})$ satisfying $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{x}_{i,n}\to 0$ and $g^{(i)}(x_{i,n})=0$. Now since each of $g^{(n)}$ is bounded (by continuity over the compact set $[-R∕2,R∕2]$, for example), each of $g^{(n)}$ also has a bounded derivative, and thus we can apply claim 2 to get that $g^{(n)}(0)=0$ for all $n$. Finally claim 3 implies that $b_n=0$ for all $n$, and hence $g(x)$ must be identically 0 over $(-R,R)$.