# Prob. 16, Chap. 5 in Baby Rudin: If $f^{\prime\prime}$ is bounded on $(0,\infty)$ and $f(x)\to 0$ as $x\to\infty$, then $f^\prime(x)\to 0$

Here is Prob. 16, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is twice-differentiable on $(0, \infty)$, $f^{\prime\prime}$ is bounded on $(0, \infty)$, and $f(x) \to 0$ as $x \to \infty$. Prove that $f^\prime(x) \to 0$ as $x \to \infty$.

Hint: Let $a \to \infty$ in Exercise 15.

And, here is Prob. 15, Chap 5 in Baby Rudin, 3rd edition:

Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a, \infty)$, and $M_0$, $M_1$, $M_2$ are the least upper bounds of $\left| f(x) \right|$, $\left| f^\prime(x) \right|$, $\left| f^{\prime\prime}(x) \right|$, respectively, on $(a, \infty)$. Prove that $$M_1^2 \leq 4 M_0 M_2.$$

Hint: If $h > 0$, Taylor's theorem shows that $$f^\prime(x) = \frac{1}{2h} \left[ f(x+2h) - f(x) \right] - h f^{\prime\prime}(\zeta)$$ for some $\zeta \in (x, x+2h)$. Hence $$\left| f^\prime(x) \right| \leq h M_2 + \frac{M_0}{h}.$$ To show that $M_1^2 = 4 M_0 M_2$ can actually happen, take $a = -1$, define $$f(x) = \begin{cases} 2x^2-1 \ & \ (-1 < x < 0), \\ \frac{x^2 - 1}{x^2 + 1} \ & \ (0 \leq x < \infty), \end{cases}$$ and show that $M_0 = 1$, $M_1 = 4$, $M_2 = 4$.

Does $M_1^2 \leq 4 M_0 M_2$ hold for vector-valued functions too?

Here is the link to an earlier post of mine on Prob. 15, Chap. 5:

Prob. 15, Chap. 5 in Baby Rudin: Prove that $M_1^2\leq M_0M_2$, where $M_0$, $M_1$, and $M_2$ are the lubs, resp., of ...

I have no idea of how to directly apply the conclusion of Prob. 15 here merely by taking $a \to \infty$. However I can come up with the following attempt:

As $f^{\prime\prime}$ is bounded on $(0, \infty)$, so we can find a positive real number $M$ such that $$\left| f^{\prime\prime}(x) \right| \leq M \ \mbox{ for all real numbers } x > 0.$$

Let $h$ be a real number such that $h > 0$. As $f(x) \to 0$ as $x \to \infty$, so, given a real number $\varepsilon > 0$, we can find a real number $\alpha$ such that $$\left| f(x) \right| < \frac{\varepsilon h}{4} \ \mbox{ for all real numbers } \ x > \alpha.$$ So, for all real $x > \alpha$, from Taylor's theorem we obtain $$f \left( x+h\right) = f(x) + h f^\prime(x) + \frac{h^2}{2!} f^{\prime\prime}(\zeta)$$ for some real number $\zeta$ such that $x < \zeta < x+h$. Therefore, $$f^\prime(x) = \frac{ f \left( x+h \right) - f(x) }{h} - h f^{\prime\prime}(\zeta).$$
So, for all real numbers $x > \alpha$ and for all real numbers $h > 0$, we note that
\begin{align} \left| f^\prime(x) \right| &= \left| \frac{ f \left( x+h \right) - f(x) }{h} - h f^{\prime\prime}(\zeta) \right| \\ &\leq \frac{ \left| f\left( x+h \right) \right| + \left| f(x) \right| }{h} + h \left| f^{\prime\prime}(x) \right| \\ &< \frac{ \frac{\varepsilon h}{4} + \frac{\varepsilon h}{4} }{h} + h \left| f^{\prime\prime}(x) \right| \\ &= \frac{\varepsilon}{2} + h \left| f^{\prime\prime}(x) \right| \\ &\leq \frac{\varepsilon}{2} + h M. \end{align} Now if we take $h$ such that $0 < h < \varepsilon/2M$, then $$\left| f^\prime(x) \right| < \varepsilon \ \mbox{ for all real numbers } x > \alpha,$$ from which it follows that $f^\prime(x) \to 0$ as $x \to \infty$, as required.

Is my proof correct? If so, then is it rigorous enough for Rudin?

• At first sight, you get the estimate $|\frac{f(x+h) - f(x)}{h}| \leq \frac{\epsilon}{2}$ using only the fact that $f(x) \to 0$ as $x\to +\infty$. If this is the case, it cannot be correct. Commented May 19, 2017 at 11:15
• @Rigel: For a fixed $h$ it's valid that $(f(x+h)-f(x))/h \to 0$, I think, although it's not true that $f'(x) \to 0$. Commented May 19, 2017 at 11:24
• For a fixed $h$ it seems ok, but you are saying that the estimate holds for every positive $h$. Maybe I'm missing some detail... Commented May 19, 2017 at 11:30
• I believe that $\lim_{h\to 0} \lim_{x\to\infty}(f(x+h)-f(x))/h = 0$ holds, but not that $\lim_{x\to\infty} \lim_{h\to 0} (f(x+h)-f(x))/h = 0$ holds. That's an important difference. Commented May 19, 2017 at 13:35

You can use Prob. 15 in the following way. Let $M_2 := \sup_{x> 0} |f''(x)|$ (so $M_2 < +\infty$ by assumption), and, for every $a> 0$, define $$M_1(a) := \sup_{x\geq a} |f'(x)|, \qquad M_0(a) := \sup_{x\geq a} |f(x)|.$$ The assumption $\lim_{x\to +\infty} f(x) = 0$ implies that $M_0(a)$ is finite for every $a> 0$ and $$\lim_{a\to +\infty} M_0(a) = 0.$$ From Prob. 15 you know that $$0 \leq M_1(a)^2 \leq 4 M_2\, M_0(a),$$ hence $M_1(a) \to 0$ as $a\to +\infty$ by comparison.
• how does the assumption $\lim_{x \to \infty} f(x) = 0$ imply that $M_0(a)$ is finite for every $a > 0$? Commented May 19, 2017 at 11:41
• Because $f$ is continuous and bounded. Commented May 19, 2017 at 11:46
• @Siminore I would be grateful if you could please give a detailed proof of why it is really so; we are only given that $f^{\prime\prime}$ is bounded, not that $f$ is bounded. Of course, $f$ is continuous on $(0, \infty)$. Commented May 19, 2017 at 16:29
• For every $a>0$, $f$ is continuous in $[a, +\infty)$. Since $f(x) \to 0$ as $x\to +\infty$, then $f$ is bounded in $[a,+\infty)$. (Proof: there exists $K>a$ such that $|f(x)| \leq 1$ for $x\geq K$; moreover, $f$ is bounded on the compact set $[a, K]$). Commented May 19, 2017 at 16:40