Behavior of $f'(x)$ as $x \to \infty$ given the behavior of $f(x)$ and $f''(x)$

Question

Suppose $f$ is twice differentiable on $(a, \infty)$ and let $\lim_{x \to \infty}f(x) = 0$. Also assume that $f''(x)$ is bounded for all $x \in (a, \infty)$. Prove that $f'(x) \to 0$ as $x \to \infty$.

Note that if $f'(x)$ tends to a limit as $x \to \infty$ then this limit must be $0$ (because of the relation $f(x) - f(x/2) = (x/2)f'(c)$). What we need to establish now is that $f'(x)$ does tend to a limit. To show that we need to somehow use the fact that $f''(x)$ is bounded.

Let $b > 0$ be any arbitrary but fixed number then we have relation $f(x + b) - f(x) = bf'(c)$ and this again implies that for all sufficiently large $x$ the derivative $f'$ takes a small value in the interval $(x, x + b)$. What we need to show that eventually all the values of $f'$ are small in such intervals.

Now we can take $b$ as small as we please, we can therefore find a value $N > 0$ such that $f'$ takes very small value for at least one point in $(x, x + b)$ for all $x > N$. If $p, q$ are two points of this interval $(x, x + b)$ then $|f'(p) - f'(q)| = |(p - q)f''(\xi)| \leq b|f''(\xi)|$ and since $f''$ is bounded it follows that $f'(p) - f'(q)$ is small so that effectively all values of $f'$ in $(x, x + b)$ are small.

The above is a non-rigorous argument which I have not been able to make rigorous by using $\epsilon, \delta$ in proper manner. Maybe the approach used above can't be made rigorous and I am wrong path (or may be not!). Please help me in making above approach rigorous or suggest some alternative method.

Further there is another generalization available:

If $f(x) $ tends to a finite limit as $x \to \infty$ and $f^{(n + 1)}(x)$ is bounded then show that $f^{(n)}(x) \to 0$ as $x \to \infty$.

I have not tried to solve this general problem and any hint would be great.

Answer 1

This can be handled with only the Mean Value Theorem.

Suppose $|f''(x)|\le M$ and $\lim\limits_{x\to\infty}f'(x)\ne0$; that is, there is an $\epsilon\gt0$ so that for any $N\gt0$ we can find an $x_N\ge N$ so that $|f'(x_N)|\ge\epsilon$.

The Mean Value Theorem says that if $|x-x_N|\le \frac\epsilon{2M}$, then for some $\xi$ between $x$ and $x_N$, $$ \begin{align} |f'(x)-f'(x_N)| &=|x-x_N||f''(\xi)|\\ &\le\frac\epsilon{2M}M\\ &=\frac\epsilon2\tag{1} \end{align} $$ Therefore, if $|x-x_N|\le \frac\epsilon{2M}$, $$ \begin{align} |f'(x)| &\ge|f'(x_N)|-|f'(x)-f'(x_N)|\\ &\ge\epsilon-\frac\epsilon2\\ &=\frac\epsilon2\tag{2} \end{align} $$ The Mean Value Theorem also says that $$ \left|f\left(x_N+\frac\epsilon{2M}\right)-f\left(x_N-\frac\epsilon{2M}\right)\right| =\frac\epsilon{M}|f'(x)|\tag{3} $$ for some $x$ where $|x-x_N|\le\frac\epsilon{2M}$. Therefore, $$ \left|f\left(x_N+\frac\epsilon{2M}\right)-f\left(x_N-\frac\epsilon{2M}\right)\right| \ge\frac{\epsilon^2}{2M}\tag{4} $$ However, if since $\lim\limits_{x\to\infty}f(x)=0$, there must be an $L$ so that if $x\ge L$, then $|f(x)|\le\frac{\epsilon^2}{6M}$. Now pick an $x_N\ge L+\frac\epsilon{2M}$ so that $f'(x_N)\ge\epsilon$. Then $$ \begin{align} \left|f\left(x_N+\frac\epsilon{2M}\right)-f\left(x_N-\frac\epsilon{2M}\right)\right| &\le\left|f\left(x_N+\frac\epsilon{2M}\right)\right|+\left|f\left(x_N-\frac\epsilon{2M}\right)\right|\\ &\le\frac{\epsilon^2}{6M}+\frac{\epsilon^2}{6M}\\ &=\frac{\epsilon^2}{3M}\tag{5} \end{align} $$ However, inequality $(4)$ assures us that the difference in $(5)$ is at least $\frac{\epsilon^2}{2M}$. Thus, our assumption that $\lim\limits_{x\to\infty}f'(x)\ne0$ is false. Thus, we must have $$ \lim_{x\to\infty}f'(x)=0\tag{6} $$

Answer 2

Write the Taylor inequality (with $|f''|\le M$) and the fact that $f(x)\to 0$ to get an $A$ such as:

$$ |f(x+h) - f(x) -hf'(x)|\le M\frac{h^2}2; \\ x\ge A \implies h|f'(x)| \le 2\epsilon + M\frac{h^2}2; \\ |f'(x)| \le \inf_{h>0} \left[2\frac{\epsilon}h + M\frac{h}2\right] =\sqrt{4M\epsilon} =O(\sqrt{\epsilon}). $$

Answer 3

Can you show $\left|\displaystyle\int_{x-\frac{|f'(x)|}K}^{x+\frac{|f'(x)|}K}f'(t)\ dt\right|\ge\frac{|f'(x)|^2}K$ ? Use the fact that if $f'$ has a bounded derivative, then $f'$ is $K$-Lipschitz (imagine a triangle above or below the function, what's its area?)

Now $\displaystyle0=\lim_{x\to\infty}\left|f\left(x+\frac{|f'(x)|}K\right)-f\left(x-\frac{|f'(x)|}K\right)\right|=\lim_{x\to\infty}\left|\int_{x-\frac{|f'(x)|}K}^{x+\frac{|f'(x)|}K}f'(t)\ dt\right|$.

Answer 4

I would have considered that a sufficiently rigorous argument! But I suppose I should require more rigor from a student just learning this stuff than a colleague who already knows this stuff.

If I may, I will summarize your argument as:

1. $f'(x)$ eventually take small values in every short short interval
2. Eventually, $f'(x)$ cannot vary much over a short interval
3. Therefore, eventually, $f'(x)$ is always small
4. $|f'(x)|$ is eventually bounded by arbitrarily small numbers, and is thus zero

I've added point 4, which I imagine you were implicitly thinking.

Let $S$ be the bound for which we call "small"; that is, let $S > 0$.

You've already rougly expressed point $1$: more symbolically, you assert we can choose $N$ and $B$ such that

For every $x > N$, there exists a $c \in (x, x+B)$ such that $|f'(c)| < S$

You've also roughly expressed point $2$. If I may slightly rephrase you assert that we can choose $N$ and $B$ so that we also have

For every $x > N$ and for every $p,q \in [x, x+B]$, we have $|f'(p) - f'(q)| < S$

The trick for point $3$ is to use the triangle inequality: your intuitive idea is that $f'(c)$ is small, and any other point in the short interval $f'(p)$ is not far from $f'(c)$, therefore $f'(p)$ is small. These "small + small = small" arguments are very frequently done by using the triangle equality to find a simpler (but looser) upper bound:

$$ |f'(p)| = \left|f'(c) + \left( f'(p) - f'(c) \right) \right| \leq |f'(c)| + | f'(p) - f'(c) < 2S $$

So what you want to show for point $3$ is, given our choice of $N$ and $B$ above:

For every $x > N$ and for every $p \in [x, x+B]$, we have $f'(p) < 2S$

But really, what you're trying to say is that all values are small: use the above to prove

For every $x > N$, $|f'(x)| < 2S$

Finally, the last piece you need to make it all work for point $4$ is to view the previous work as proving:

For every $S > 0$ we can choose an $N$ such that for all $x > N$: $|f'(x)| < 2S$

No need to do anything special at this point: your previous work is pretty much already in the form you need! You just need to organize it and clearly state what you've shown. And if you're nitpicky, do one last final step using this to prove an $\epsilon-N$ argument.

I've left the blockquoted sections for you to prove; I hope that having a clearly stated target will make it easier to figure out what you're doing. (and that this will help you organize your thoughts in the future!)

Answer 5

Using mookid's idea of using Taylor's series I have come up with a nice solution to the general problem. Also if $f(x) \to L$ as $x \to \infty$ then we can consider the function $g(x) = f(x) - L$ and hence without any loss of generality we can assume that $L = 0$.

Let $\epsilon > 0$ be arbitrary. Let $K > 0$ be a bound for $|f^{(n + 1)}|$. Also let $h > 0$ be a number to be chosen based on $\epsilon$ and $K$ (exact nature of dependence of $h$ on these two numbers will be seen later). Consider the sum $$F(x, h) = \sum_{i = 0}^{n}(-1)^{n - i}\binom{n}{i}f(x + ih)$$ If we put $\Delta x$ in place of $h$ it is easy to seen that the expression $F(x, h) = F(x, \Delta x)$ is nothing but the repeated difference $\Delta^{n}f(x)$. It can be shown that there is value $p \in (0, nh)$ such that $$F(x, h) = h^{n}f^{(n)}(x + p)$$ (this could be called another form of Taylor's theorem). Since $f(x) \to 0$ as $x \to \infty$, there is a number $N > 0$ corresponding to $\epsilon > 0$ such that $|f(x)| < \epsilon$ for $x > N$. From the formula for $F(x, h)$ and noting that $h > 0$ we can see that $|F(x, h)| \leq M\epsilon$ where $M$ is finite constant dependent on $n$. Next we can consider the values of $f^{(n)}$ in interval $[x, x + nh]$. Clearly for any $\xi \in [x, x + nh]$ we have $|f^{(n)}(\xi) - f^{(n)}(x + p)| \leq nhf^{(n + 1)}(c) \leq nh K$. It now follows that $$|f^{(n)}(\xi)| \leq |f^{(n)}(x + p)| + nhK \leq \frac{M\epsilon}{h^{n}} + nhK$$ By maxima/minima technique we can get a positive value of $h$ which minimizes the right side. This value of $h$ is given by $h^{n + 1} = M\epsilon/K$ and the minimum value of the expression on the right is of the form $O(\epsilon^{1/(n + 1)})$.

It now follows that for any given value $\epsilon > 0$ there is a value $N > 0$ and an $h > 0$ such that for any $x > N$ the values of $f^{(n)}$ in the interval $[x, x + nh]$ are of the form $O(\epsilon^{1/(n + 1)})$. Since $nh > 0$ and then $x + nh > N$ so we can change the interval to $[x + nh, x + 2nh]$ and result remains same. It follows by induction that the values of $f^{(n)}(x)$ are bounded by a term of type $O(\epsilon^{1/(n + 1)})$ for all $x > N$. It follows that $f^{(n)}(x) \to 0$ as $x \to \infty$.