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Let $f \in C^\infty([0,+\infty),\mathbb{R})$, and $f(0)=\lim\limits_{x \to \infty}f(x)$

Prove that there exist $c_n>0$ such that $f^{(n)}(c_n)=0 $ for all $n$ integer, where $f^{(n)}$ is the n-th derivative of $f$

My attempt:

If $f$ is constant then the result is obvious.

If not there exist $x_0 \in (0,+\infty)$ such that $f(x_0) \neq f(0)$

Let $y=\frac{1}2 (f(x_0)+f(0))$, then by IVT theorem there exist $a \in (0,x_0)$ such that $f(a)=y$

Despite $f(0)=\lim_{+\infty}f(x)$, $y$ is an intermediate value of $f(x_0)$ and $f(x_1)$ with $x_1$ sufficiently large.

Then, by IVT there exist $b \in (x_0,x_1]$ such that $f(b)=y$

Therefore by Rolle's theorem on $[a,b]$ we can conclude that $f'$ vanishes.

Unfortunately I don't see any direct proof to show that $f^{(n)}$ vanishes.

I managed to find a HINT in a book :

Prove that either $\lim\limits_{x \to \infty}f'(x)=0$ or $f'$ admits an infinite number of extrema.

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  • $\begingroup$ I don't understand the question. Do you want to show there is a single $c>0$ such that $f^{(n)}(c) = 0$ for all $n$? Is there a particular $n$ you would like to look at? Can $c$ change depending on $n$? Also, some of your notation is weird: Should that be $\lim_{x\to \infty}$? And what is $a >\in (0,x_0)$? $\endgroup$ – Tyler Holden Mar 5 '14 at 16:39
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    $\begingroup$ You have badly mis-stated the question! In general, there won't be a single $c$ for which $f^n(c) = 0$ for all $n$. Please edit accordingly. $\endgroup$ – TonyK Mar 5 '14 at 20:25
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You have found the solution for the first derivative, the idea is then to use this solution to bootstrap yourself to the next derivative.

First of all, you may assume that $f(0) =0$, since you may always translate the function vertically without affecting zeroes of the derivatives. Thus the question can be thought of equivalently as:

If $\displaystyle\lim_{x\to \infty} f(x) = f(0) = 0$ show that for every $n \in \mathbb N$, there exists some $c_n$ such that $f^{(n)}(c_n)=0$.

As motivation, consider the following incorrect but useful idea: It is tempting to say that if $f(x)$ converges as $x\to \infty$ then $f'(x) \xrightarrow{x\to\infty} 0$, though this is not true! (For example, $\frac1x\sin(x^2)$ is $C^\infty$ on $[0,\infty)$.) Nonetheless, let's think about what would happen if this were true. By your argument above, you know that there is some $c_1$ such that $f'(c_1) =0$ and $\displaystyle\lim_{x\to\infty} f'(x) = 0$. This is precisely the same question, only now you have shifted the problem from $0$ to $c$! Hence your same argument above shows there is some $c_2$ such that $f''(c_2) = 0$. Now proceed inductively to get the remaining derivatives.

The problem of course is dealing with the fact that $\lim_{x\to \infty} f(x)$ need not converge to zero (or converge at all for that matter). Heuristically, this will happen if the derivative is constantly oscillating quicker and quicker as $x$ gets larger, but such functions typically exhibit the desired qualities of the problem anyway! So how do we circumnavigate this problem? We take a discrete sampling and use the same argument. Hence one way you might proceed is as follows:

First note that clearly $f(x)$ and $f(y)$ get arbitrarily close for sufficiently large $x$ and $y$. Indeed, by definition we know that for all $\epsilon >0$ there is an $N \in \mathbb R$ such that $x>N$ implies $|f(x)-f(0)|<\frac\epsilon2$. Hence if $x,y>N$ then $$|f(x)-f(y)| \leq |f(x)-f(0)|+|f(y)-f(0)| < \epsilon.$$ Let $I_n = [2n, 2n+1]$. By the Mean Value Theorem, there is some $d_n$ such that $f'(d_n) = f(2n+1)-f(2n)$, and as $n \to \infty$, so that $f'(d_n) \xrightarrow{n\to\infty}0$ by the above argument. Notice that this is essentially what we wanted to say in my motivational paragraph, only that it wasn't true with a continuous parameter $x$, but is now true using a sequence of $(d_n)_{n=1}^\infty$.

Now your argument above shows that there is some $c$ such that $f'(c) =0$. By the extreme value theorem, $f'(x)$ will always take its max/min on the interval $[c,d_n]$. It is easy to argue that since $f'(d_n) \to 0$, then for sufficiently large $n$, the max/min occurs on the interior of $[c,d_n]$, say at the point $c_2$. This point $c_2$ is critical for $f'(x)$, and so satisfies $f''(c_2) =0$.

Now bootstrap: You want to find a sequence of $d_n^2$ such that $f''(d_n^2) \to 0$. By the MVT, there is a point $d_n^2 \in (d_{2n},d_{2n+1})$ such that $$f''(d_n^2) = \frac{f'(d_{2n+1})-f'(d_{2n})}{d_{2n+1}-d_{2n}} \leq f'(d_{2n+1})-f'(d_{2n})$$ (since $d_{2n+1}-d_{2n}>1$: think about this), and this goes to zero. Rinse and repeat.

Edit: Hopefully it is clear that your hint is embedded in my solution. Again, if you have that $\displaystyle \lim_{x\to\infty} f'(x) = 0$ and $f'(c_1)=0$ then you are done since you just repeat your argument to find a $c_2 > c_1$ such that $f''(c_2) = 0$. If $f'(x)$ admits infinitely many extrema (this is the case I circumvented by taking a discrete sampling) then one of those extrema lies on the interior (this is my $[c,d_n]$) argument, and so the derivative has a point which vanishes.

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