Proving $f : (-1, 1) \rightarrow \mathbb{R}$ has a property

Solution 1:

For the second question: Suppose the result fails. Then $f^{(k)}(0)\ne 0$ for some $k\in \{0,1,\dots,n-1\}.$ Let $k_0$ be the smallest such $k.$ We then have $f^{(k_0)}(0)\ne 0$ and $f^{(k)}(0)= 0$ for $k<k_0.$

By the theorem you cited, for each $x\in (-1,1)\setminus\{0\}$ there exists $z_x$ beween $0$ and $x$ such that

$$f(x) = \frac{f^{(k_0)}(z_x)}{k_0!}x^{k_0}.$$

It follows that for such $x.$

$$M|x|^n \ge \frac{|f^{(k_0)}(z_x)|}{k_0!}|x|^{k_0}.$$

Therefore

$$Mk_0!|x|^{n-k_0} \ge |f^{(k_0)}(z_x)|.$$

Now $n>k_0.$ Thus as $x\to 0,$ the left side $\to 0.$ Because $x\to 0$ forces $z_x\to 0,$ the right side $\to |f^{(k_0)}(0)|.$ The reason I can say the last is that $f^{(k_0)}$ is differentiable on $(-1,1),$ hence is continuous there.

We have thus shown $f^{(k_0)}(0)=0,$ contradiction. Thus there is no such $k_0,$ and this gives the desired conclusion.

Solution 2:

The theorem mentioned in your post is Taylor's theorem with Lagrange's form of remainder stated in a peculiar form. And your solution to the first problem based on this theorem is fine. +1 to your question for that.

Note however that both the problems can be easily solved using the much simpler mean value theorem and the slightly complicated L'Hospital's Rule. Let's attack the first problem. If $x\in(-1,1)$ then $$f^{(n-1)}(x)=f^{(n-1)}(x)-f^{(n-1)}(0)=xf^{(n)}(z)$$ and hence by given hypothesis we have $$|f^{(n-1)}(x)|\leq M|x|$$ Next we have in similar manner $$|f^{(n-2)}(x)|=|xf^{(n-1)}(z')|\leq M|x||z'|<M|x|^2$$ as $z'$ lies between $0$ and $x$. Continuing in this fashion we ultimately get $|f(x) |\leq M|x|^n$.

The second problem asks us to do the reverse. And for first derivative this is nothing more than an application of definition of derivative. Since $|f(x) |\leq M|x|^n$ we have $f(0)=0$ and then $$0\leq \left|\frac{f(x)} {x} \right|\leq M|x|^{n-1}$$ By squeeze theorem $f(x) /x\to 0$ as $x\to 0$. Thus $f'(0)=0$. Evaluation of further derivatives requires the use of L'Hospital's Rule. In general it can be proved using L'Hospital's Rule that if $f^{(n)}(0)$ exists then $$f^{(n)} (0)=\lim_{x\to 0}\frac{n! }{x^n}\left(f(x)-f(0)-xf'(0)-\frac{x^2}{2!}f''(0)-\dots-\frac{x^{n-1}}{(n-1)!}f^{(n-1)}(0)\right)$$ Using this formula we can show that $f^{(k)} (0)=0$ for $k=2,3,\dots,n-1$. For example consider $$f(x) - f(0)-xf'(0)=f(x)$$ and hence $$0\leq\frac{2!}{|x|^2}\left|f(x)-f(0)-xf'(0)\right|\leq 2!M|x|^{n-2}$$ and by squeeze theorem we have $$f''(0)=\lim_{x\to 0}\frac{2!}{x^2}\left(f(x)-f(0)-xf'(0)\right)=0$$

You can also use theorem of your question to solve the second problem. But you need a trick similar to the one used in solution presented above. If $$g(x) =f(x) - \sum_{k=0}^{n-1}\frac{x^k}{k!}f^{(k)}(0)$$ then $$g(0)=g'(0)=\dots=g^{(n-1)}(0)=0,g^{(n)}(x) =f^{(n)} (x) $$ and by theorem in question we have $$g(x) =\frac{x^n} {n!} f^{(n)} (z) $$ where $z$ lies between $0$ and $x$. And this brings us to the standard Taylor's theorem with Lagrange's form of remainder $$f(x) =\sum_{k=0}^{n-1}\frac{x^k}{k!}f^{(k)}(0)+\frac{x^n}{n!}f^{(n)}(z)$$ Now replace $n$ by $1,2,\dots,n-1$ and use the fact that $|f(x) |\leq M|x|^n$ to conclude that $f^{(k)} (0)=0$ for $k=1,2,\dots,n-1$.