I am interested in finding Example of twice differentiable function on $(0,\infty)$ $f(x)\to 0 , x\to \infty$ but $f'(x)$ not tends to 0 as $x$ goes to infinity.
I had already proved that in case of $f''(x)$ is bounded then we can show that above is not true. I tried to get some examples, but did not succeed. Any help will be appreciated!
Best Answer
The idea to approach a question like yours, is that $f$ must oscillate heavily as we approach the tail, but the amplitude must be steadily decreasing to zero. Essentially, try to make $f$ a product of two terms : one curbing the oscillation, ensuring the function goes to zero, and the other increasing the frequency of oscillation (keeping amplitude constant), ensuring the derivative goes to infinity.
For example, taking $\frac 1x$, which goes to infinity, and $\sin (x^2)$, which oscillates rapidly, satisfy these conditions (note that $\sin x$ also oscillates, but $\frac 1x$ doesn't just go to infinity : it also provides a dampening effect to oscillations, since its derivative is $x$ to a lower power. To counter this lower power, we need more rapid oscillations). You can check that $\frac 1x \sin(x^2)$ is a counterexample. So is $\frac 1x \sin x^3$.
For fun, try to come up with conditions on $f$ and $g$ so that $h(x) = f(x) \sin g(x)$ is a counterexample to your assertion.
As you have noted , $f''$ being bounded implies that if $f(x) \to 0$ then $f'(x) \to 0$ as $x \to \infty$. However, there is a more general condition, which is weaker than $f''$ being bounded. It is the uniform continuity of $f'$ : if $f'$ is assumed uniformly continuous, rather than differentiable with bounded derivative, then too it is true that it would converge to $0$.You can try this as an exercise.
Answer to the exercise
Known as Barbalat's lemma, is the statement that if $f$ is differentiable on $(a,\infty)$, continuous on $[a,\infty)$ , and $f'$ is uniformly continuous on $(a,\infty)$, then $\lim_{x \to \infty} f(x) = a < \infty$ implies that $\lim_{x \to \infty} f'(x) = 0$. Note that we actually have $a= 0$, but the value of $a$ does not matter, because one may add to , or subtract a constant from $f$ to make the value of $\lim_{x \to \infty}f(x)$ change : the derivative removes constants, so it will not change.
We prove this by contradiction. Suppose that $\lim_{x\ to \infty} f'(x) \neq 0$.We negate the definition of limit equalling zero, to get : there exists $\epsilon > 0$, for all $r$ there exists $x > r$ such that $|f'(x)| > \epsilon$.
Now, take $r = 1,2,...$ in this statement, to get points $x_i$ at which $|f'| > \epsilon$. There will be infinitely many such points. This implies that either the set of points for which $f' < -\epsilon$, or the set of points for which $f' > \epsilon$,(or both) will be an infinite set. Without loss of generality, let us assume that $f'(x_i) > \epsilon$ for all $i$.
Now, we will see what happens if $f'$ is just continuous.
Now, what happens under uniform continuity?
Ok, so what extra is uniform continuity giving us? Not clear so far.
Consider the quantity $D_i = \int_{a}^{x_i + \delta} f' - \int_{a}^{x_i} f' = \int_{x_i}^{x_i + \delta} f'$. Since $f' > \epsilon$ on these intervals, we see that $D_i > \epsilon \delta$ for all $i$.
However, from the fundamental theorem of calculus, we know that $D_i = f(x_i + \delta) - f(x_i)$. Therefore, since $\lim_{x \to \infty} f(x) = \alpha$, the limit of the RHS exists and equals zero. Consequently , $\lim_{i \to \infty} D_i = 0$. But this can't happen : $D_i > \epsilon \delta$, so it can't get closer than this to zero! Contradiction.
What happens if we change back to just continuity? The problem is that $D_i$ is now $\int_{x_i}^{x_i + \delta_i} f'$. The bound created is $D_i >\epsilon\delta_i$. Since $\delta_i$ is not fixed, this does not prevent $D_i$ from converging to zero! Which was the trick to producing the contradiction.