Jtdylktuy

I am interested in finding Example of twice differentiable function on $(0,infty)$ $f(x)to 0 , xto infty$ but $f'(x)$ not tends to 0 as X goes to infinity.
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I had already proved that in case of f'' is bounded then we can show that above is not true. I tried to get some examples.but not succeed. Any help will be appreciated

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.

Since $f'$ is continuous at each $x_i$, there exists $delta_i$ depending on $x_i$, such that $|y - x_i| < delta_i$ implies $f'(y) > epsilon$.

Now, what happens under uniform continuity?

Since $f'$ is uniformly continuous, there exists $delta$ not depending on $x_i$, such that $|y - x_i| < delta$ implies $f'(y) > epsilon$.

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 $D_i >epsilondelta_i$. Since $delta_i$ is not fixed, this does not prevent $D_i$ from converging to zero! Which was the trick to producing the contradiction.