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

Question

I'd like to answer the two questions below. I believe my solution to the first one is right, but it'd help if someone could please help verify this, as I am working through this book independently through self-study. I don't know how to do the second one, which is a continuation of the first problem. I'm pretty sure both questions utilize a Theorem stated in my book, so I've provided that as well.

Suppose $f : (-1, 1) \rightarrow \mathbb{R}$ has $n$ derivatives and its $n^{\text{th}}$, suppose its $n^{\text{th}}$ derivative $f^{(n)} : (-1, 1) \rightarrow \mathbb{R}$ is bounded. Finally, suppose we have

$$f(0) = f'(0) = f''(0) = \cdots f^{(n - 1)}(0) = 0.$$

Prove there exists a positive number $M$ such that $$|f(x)| \leq M|x|^{n}.$$

Second question:

Suppose $f : (-1, 1) \rightarrow \mathbb{R}$ has $n$ derivatives and there is a positive number $M$ such that $$|f(x)| \leq M|x|^{n}.$$ Prove $$f(0) = f'(0) = \cdots f^{(n - 1)}(0) = 0.$$

Also, here's the theorem I was referring to:

Theorem: Let $I$ be an open interval and $n$ be a natural number and suppose that the function $f : I \rightarrow \mathbb{R}$ has $n$ derivatives. Suppose also at the point $x_{0}$ in $I$, $$f^{(k)}(x_{0}) = 0$$ for $0 \leq k \leq n - 1$. Then, for each point $x \neq x_{0}$ in $I$, there is a point $z$ strictly between $x$ and $x_{0}$ at which $$f(x) = \frac{f^{(n)}(z)}{n!}(x - x_{0})^{n}.$$

---

Here's my attempt at problem $1$:

By the Theorem above, for each $x \in (-1, 1)$, $x \neq 0$, there's a point $z$ between $x$ and $0$ such that $$f(x) = \frac{f^{(n)}(z)}{n!}x^{n}.$$

Using the fact that $f^{(n)}$ is bounded, we know there exists a bound $N$ such that $|f^{(n)}(x)| \leq N$ for all $x$ in $(-1, 1)$. Therefore,

$$|f(x)| = \left|\frac{f^{(n)}(z)}{n!}x^{n}\right| \leq |\frac{N}{n!}x^{n}| = (N/n!)|x|^{n} $$

So setting $M = N/n!$ suffices.

Is this correct? How do I do the next one?

Answer 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.

Answer 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 $k=1,2,\dots,n-1$.

Answer 3

Paramanand's answer gave me an idea:

It is a fact that $$(-1)^nf^{(n)}(x)=\lim_{h\rightarrow 0}\frac{1}{h^n}\sum_{k=0}^n(-1)^k\binom{n}{k}f(x+kh)$$ if $f$ is $n$-times differentiable at $x$ (with $f^{(n)}$ continuous at $x$).

This is, essentially, Newton forward differences.

Let's now fix a $k$ between $1$ and $n-1$ (this claim is clear when $k=0$). Since $f$ is $n$-times differentiable, we know that $f^{(k)}$ is continuous for $1\leq k\leq n-1$.

We have that $$|f^{(k)}(0)|\leq \lim_{h\rightarrow 0}\frac{1}{|h|^k}\sum_{j=0}^k\binom{k}{j}|f(jh)|\leq \lim_{h\rightarrow 0}|h|^{n-k}\sum_{j=0}^k\binom{k}{j}M|j|^n=0\,.$$

An algebraic proof to the identity I've used is given here.

A simple analytic proof is given here.