# 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 …

Here is Prob. 15, Chap. 5 in the book Principles of Mathematical Analysis by Walter 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 $M_1^2 = 4M_0M_2$ 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?

My Attempt:

Let $x > a$ and $h > 0$. Then by Taylor's theorem (i.e. Theorem 5.15 in Baby Rudin, 3rd edition), we can find a real number $\zeta \in (x, x+2h)$ such that $$f(x+2h) = f(x) + 2h f^\prime(x) + \frac{(2h)^2}{2!} f^{\prime\prime}(\zeta) = f(x) + 2h f^\prime(x) + 2h^2 f^{\prime\prime}(\zeta),$$ and solving for $f^\prime(x)$ we obtain $$f^\prime(x) = \frac{1}{2h} \left[ f(x+2h) - f(x) \right] - h f^{\prime\prime}(\zeta),$$ and hence \begin{align} \left| f^\prime(x) \right| &\leq \frac{1}{2h} \left[ \left| f(x+2h) \right| + \left| f(x) \right| \right] + h \left| f^{\prime\prime}(\zeta) \right| \\ &\leq \frac{1}{2h} \left( M_0 + M_0 \right) + h M_2 \\ &= \frac{ M_0}{h} + h M_2. \end{align} Thus for all $x \in (a, \infty)$ and for all $h > 0$, we see that $$\left| f^\prime(x) \right| \leq M_2 h + \frac{M_0}{h}.$$ Now keeping $h$ fixed and taking the supremum of the left-hand side over all $x \in (a, \infty)$, we obtain $$M_1 \leq h M_2 + \frac{M_0}{h} \ \mbox{ for all h > 0. } \tag{1}$$ Now if $M_2 = 0$, then (1) yields $$0 \leq M_1 \leq \frac{M_0}{h} \ \mbox{ for all h > 0. } \tag{2}$$ But $$\lim_{h \to \infty} \frac{M_0}{h} = 0;$$ so from (2) we can conclude that $M_1 = 0$ also and hence $$M_1^2 = 0 = 4M_0 M_2,$$ as required. Am I right?

So let us suppose that $M_2 > 0$, and let us define a function $\varphi$ on $(0, \infty)$ by the formula $$\varphi(h) \colon= h M_2 + \frac{M_0}{h} \ \mbox{ for all h > 0. }$$ Then $$\varphi^\prime(h) = M_2 - \frac{M_0}{h^2} = \frac{M_2}{h^2} \left( h^2 - \frac{M_0}{M_2} \right) \ \mbox{ for all h > 0. }$$ Thus, $$\varphi^\prime(h) \begin{cases} > 0 \ & \ \mbox{ if } 0 < h < \sqrt{\frac{M_0}{M_2}}, \\ = 0 \ & \ \mbox{ if } h = \sqrt{\frac{M_0}{M_2}}, \\ < 0 \ & \ \mbox{ if } h > \sqrt{\frac{M_0}{M_2}}. \end{cases}$$ Thus by Theorem 5.11 in Baby Rudin, our $\varphi$ is monotonically (in fact, strictly) decreasing for $0 < h \leq \sqrt{\frac{M_0}{M_2}}$ and monotonically (in fact strictly) increasing for $h \geq \sqrt{\frac{M_0}{M_2}}$. Therefore $\varphi$ has a local minimum at $h = \frac{M_0}{M_2}$. Moreover, since $$\varphi \left(\sqrt{\frac{M_0}{M_2}} \right) = M_2 \sqrt{\frac{M_0}{M_2}} + \frac{M_0}{\sqrt{\frac{M_0}{M_2}}} = 2 \sqrt{M_0 M_2}$$ is the only local extreme value that $\varphi$ has, so it is in fact the absolute minimum value of $\varphi$. Therefore we can conclude that $$\varphi \left( \sqrt{ \frac{M_0}{M_2} } \right) = 2 \sqrt{M_0 M_2} \leq \varphi(h) \ \mbox{ for all } h > 0. \tag{3}$$ And, since $\sqrt{\frac{M_0}{M_2}} > 0$, therefore from (1) and (3) we can also conclude that $$0 \leq M_1 \leq 2 \sqrt{M_0 M_2},$$ which (upon squaring both sides) yields $$M_1^2 \leq 4 M_0 M_2,$$ as required. Am I right?

Now let $f$ be defined on $(-1, \infty)$ by the formula $$f(x) = \begin{cases} 2x^2-1 \ & \ \mbox{ if } -1 < x < 0, \\ \frac{x^2-1}{x^2+1} = 1 - \frac{2}{x^2+1} \ & \ \mbox{ if } 0 \leq x < \infty. \end{cases}$$ Then $$f^\prime(x) = \begin{cases} 4x \ & \ \mbox{ if } -1 < x < 0, \\ 0 \ & \ \mbox{ if } x = 0, \\ \frac{4x}{(x^2+1)^2} \ & \ \mbox{ if } 0 < x < \infty. \end{cases} \tag{*}$$ At $x= 0$, we see that $$\lim_{h \to 0+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0+} \frac{ \frac{h^2-1}{h^2+1} - (-1) }{h} = \lim_{h \to 0+} \frac{2h}{h^2+1} = 0,$$ and $$\lim_{h \to 0-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0-} \frac{ 2h^2-1 - (-1)}{h} = \lim_{h \to 0-} 2h = 0.$$ Therefore $f^\prime(0) = 0$.

As $f^\prime$ is negative in $(-1, 0)$, zero at $x = 0$, and positive on $(0, \infty)$, so $f$ is (strictly) decreasing on $(-1, 0]$ and (strictly) increasing on $[0, \infty)$. Therefore $f$ has a local minimum value at $x=0$, and $f(0) = -1$, being the unique extreme value, is also the absolute minimum value of $f$. Thus $$f(x) \geq -1 = f(0) \ \mbox{ for all } x \in (-1, \infty). \tag{4}$$

On the other hand, we note that $f$ is strictly increasing on $[0, \infty)$, and $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2-1}{x^2+1} = \lim_{x \to \infty} \frac{ 1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} = \frac{1-0}{1+0} = 1.$$ So we can conclude that $$f(x) < 1 \ \mbox{ for all } x \in [ 0, \infty) \tag{5}$$ And, as $f$ is strictly decreasing on $(-1, 0]$ and as $$\lim_{x \to -1+} f(x) = \lim_{x \to -1+} \left( 2x^2-1 \right) = 1,$$ so we can conclude that $$f(x) < 1 \ \mbox{ for all } x \in (-1, 0]. \tag{6}$$ From (4), (5), and (6), we can conclude that $$\left| f(x) \right\vert \leq 1 = \left| f(0) \right| = \left| f(0) \right| \ \mbox{ for all } x \in (-1, \infty). \tag{7}$$ Thus we have $M_0 = 1$.

From (*) above, we find that $$f^{\prime\prime}(x) = \begin{cases} 4 \ & \ \mbox{ if } -1 < x \leq 0, \\ \frac{4}{(x^2+1)^2} - \frac{16x^2}{(x^2+1)^3} = \frac{4\left( 1-3x^2 \right) }{(x^2+1)^3} \ & \ \mbox{ if } 0 < x < \infty. \end{cases} \tag{**}$$ At $x = 0$, we find that $$\lim_{h \to 0+} \frac{ f^\prime(h) - f^\prime(0)}{h} = \lim_{h \to 0+} \frac{4}{(h^2+1)^2} = 4,$$ and $$\lim_{h \to 0-} \frac{ f^\prime(h) - f^\prime(0) }{h} = \lim_{h\to 0-} \frac{4h}{h} = 4,$$ and hence $f^{\prime\prime}(0) = 4$.

Now from (**) we see that $$f^{\prime\prime}(x) \mbox{ is } \begin{cases} \mbox{ positive } \ & \mbox{ if } -1 < x < \frac{1}{\sqrt{3}}, \\ \mbox{ zero } \ & \mbox{ if } x = \frac{1}{ \sqrt{3}}, \\ \mbox{ negative } \ & \mbox{ if } \frac{1}{\sqrt{3}} < x < \infty. \end{cases}$$ Thus, $f^\prime$ is strictly increasing on $\left(-1, \frac{1}{\sqrt{3}} \right]$ and strictly decreasing on $\left[ \frac{1}{\sqrt{3}}, \infty \right)$. So $f^\prime$ has a local maximum value at $x = \frac{1}{\sqrt{3}}$, which, being the only extreme value of $f^\prime$, is also the absolute maximum value. Thus, $$f^\prime \left( \frac{1}{\sqrt{3}}\right) = \frac{ \frac{4}{\sqrt{3}} }{ \left( \frac{1}{3} + 1 \right)^2 } = \frac{ \frac{4}{\sqrt{3}} }{ \frac{16}{9} } = \frac{3 \sqrt{3}}{4},$$ and $$f^\prime(x) \leq f^\prime \left( \frac{1}{\sqrt{3}}\right) = \frac{3 \sqrt{3}}{4} \ \mbox{ for all } x \in (-1, \infty). \tag{8}$$ From (*) we also note that $f^\prime(x) \to -4$ as $x \to -1+$, which together with (8) implies that $$-4 < f^\prime(x) \leq \frac{3\sqrt{3}}{4} \ \mbox{ for all } x \in (-1, \infty),$$ and so $\left| f^\prime(x) \right| \to 4$ as $x \to -1+$, which implies that our $M_1 = 4$. Am I right?

Now form (**) we see that $$f^{(3)}(x) = \begin{cases} 0 \ & \ \mbox{ if } -1 < x \leq 0, \\ -\frac{24x}{(x^2+1)^3} + \frac{24x(3x^2-1) }{((x^2+1)^4 } = \frac{ 48x \left( x^2-1 \right)}{(x^2+1)^4} \ & \mbox{ if } 0 < x < \infty. \end{cases} \tag{***}$$ At $x = 0$, we find that $$\lim_{h \to 0-} \frac{ f^{\prime\prime}(h) - f^{\prime\prime}(0)}{h} = \lim_{h \to 0-} \frac{4-4}{h} = 0,$$ and $$\lim_{h \to 0+} \frac{ f^{\prime\prime}(h) - f^{\prime\prime}(0) }{h} = 4 \lim_{h \to 0+} \frac{ 1- 3h^2 - (h^2+1)^3 }{ h (h^2+1)^3} = 4 \lim_{h \to 0+} \frac{h^5-3 h^3- 6h }{ (h^2+1)^3} = 0,$$ showing that $f^{(3)}(0) = 0$.

Thus we find that $$f^{(3)}(x) \ \mbox{ is } \begin{cases} \mbox{ negative } \ & \mbox{ if } 0 < x < 1, \\ \mbox{ zero } \ & \mbox{ if } -1 < x \leq 0 \ \mbox{ or if } x =1, \\ \mbox{ positive } \ & \mbox{ if } 1 < x < \infty. \end{cases}$$ Thus $f^{\prime\prime}$ is strictly decreasing on $[0, 1]$ and strictly increasing on $[1, \infty)$. So $f^{\prime\prime}$ has a local minimum at $x=1$, and $f^{\prime\prime}(1) = -1$, and moreover $f^{\prime\prime}(x) \to 0$ as $x \to \infty$. So $$0 \leq \left| f^{\prime\prime}(x) \right| \leq 4,$$ and $4 = f^{\prime\prime}(0)$. So our $M_2 = 4$, as required. Am I right?

Is whatever I've done so far correct? Is my approach correct? And if so, then is my solution correct and lucid enough in its presentation?

What is the situation for vector-valued functions?

• could be streamlined a bit but otherwise this looks good (+1) – tired May 14 '17 at 22:30
• Wow you typed a lot – user217285 May 14 '17 at 23:02
• For vector valued function, you might consider the function defined by $g(x)=f(x)\cdot c$, where $c$ is a constant vector, then apply the result to $g$, and use Cauchy-Schwartz inequality. – Li Chun Min May 14 '17 at 23:35
• @tired can you please point out exactly where my proof is in need of streamlining? I'd be grateful. – Saaqib Mahmood May 15 '17 at 13:30