Solutions To Mathematics Textbooks/Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)/Chapter 5

Chapter 5

6

We will compute the derivative and show that it is always $g'(x)\geq 0$. The derivative of $g(x)$ is given by the quotient rule to be $g'(x)={\frac {xf'(x)-f(x)}{x^{2}}}$ The denominator is clearly always positive as we know it is the square of some real number $x>0$, so in order to show that the entire derivative is always positive we need to show that the numerator is positive.

However, ${\frac {f(x)}{x}}\leq f'(x)\Leftrightarrow xf'(x)-f(x)\geq 0$ and we know that, by the mean value theorem there is some $c\in (0,x)$ such that ${\frac {f(x)}{x}}=f'(x)$ since $f(0)=0$. Since $f$ is monotonically increasing it must be that $f'(x)\geq {\frac {f(x)}{x}}$.

Thus, $xf'(x)-f(x)\geq 0$ and so the derivative of $g$ is positive, which implies that $g$ is monotonically increasing.

11

Since the denominator of $\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}$ goes to zero as $h\to 0$ we can employ L'Hopital's rule. Using this we get $\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}=\lim _{h\to 0}{\frac {f'(x+h)-f'(x-h)}{2h}}$ Since the denominator is again zero we apply L'Hopital's rule again $=\lim _{h\to 0}{\frac {f''(x+h)+f''(x-h)}{2}}$ $=f''(x)$

An example of a function that is not differentiable, but where this limit exists is $f(x)=\left\{{\begin{array}{lr}0&x\leq 0\\x^{2}&x>0\end{array}}\right.$ The second derivative of this function does not exist at zero, as $f'(x)=\left\{{\begin{array}{lr}0&x\leq 0\\2x&x>0\end{array}}\right.$ however we can compute the above limit at zero by $\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}=\lim _{h\to 0}{\frac {h^{2}}{h^{2}}}=1$

14

A function $f$ is said to be convex if $f(\lambda +(1-\lambda )y)\leq \lambda f(x)+(1-\lambda )f(y)$

First we shall prove ($\Rightarrow$) that $f'$ monotonically increasing implies $f$ is convex.

Assume $f'$ is monotonically increasing. Let $x<y<z$ for $x,y,z\in (a,b)$. The mean value theorem implies that $\exists c\in (x,y)$ such that $f'(c)={\frac {f(y)-f(x)}{y-x}}$. Similarly $\exists d\in (y,z)$ such that $f'(d)={\frac {f(z)-f(y)}{z-y}}$. Thus, since $f'$ is assumed to be monotonically increasing we have $f'(c)\leq f'(d)\Rightarrow {\frac {f(y)-f(x)}{y-x}}\leq {\frac {f(z)-f(y)}{z-y}}$ Because $y\in (x,z)$ we know that $y=\lambda x+(1-\lambda )z$ for some $\lambda \in (0,1)$. Moreover, we have ${\frac {f(y)-f(x)}{y-x}}\leq {\frac {f(z)-f(y)}{z-y}}$ $\Rightarrow {\frac {f(\lambda x+(1-\lambda )z)-f(x)}{\lambda x+(1-\lambda )z-x}}\leq {\frac {f(z)-f(\lambda x+(1-\lambda )z)}{z-\lambda x-(1-\lambda )z}}$ $\Rightarrow {\frac {f(\lambda x+z-z\lambda )-f(x)}{\lambda x+z-\lambda z-x}}\leq {\frac {f(z)-f(\lambda x+z-\lambda z)}{z-\lambda x-z+\lambda z}}$ $\Rightarrow {\frac {f(\lambda x+z-z\lambda )-f(x)}{\lambda x+z-\lambda z-x}}\leq {\frac {f(z)-f(\lambda x+z-\lambda z)}{-\lambda x+\lambda z}}$ $\Rightarrow \lambda (z-x)\left(f(\lambda x+z-z\lambda )-f(x)\right)\leq \left(f(z)-f(\lambda x+z-\lambda z)\right)\left(1-\lambda \right)\left(z-x\right)$ We note here that the direction of the inequality is preserved since we have $x<z$ and $\lambda >0$ and $1-\lambda >0$ (*) $\Rightarrow \lambda \left(f(\lambda x+z-z\lambda )-f(x)\right)\leq \left(f(z)-f(\lambda x+z-\lambda z)\right)\left(1-\lambda \right)$ $\Rightarrow \lambda \left(f(\lambda x+z(1-\lambda ))-f(x)\right)\leq \left(f(z)-f(\lambda x+z(1-\lambda ))\right)\left(1-\lambda \right)$ $\Rightarrow \lambda f(\lambda x+z(1-\lambda ))-\lambda f(x)\leq f(z)-f(\lambda x+z(1-\lambda ))-\lambda f(z)+\lambda f(\lambda x+z(1-\lambda ))$ reducing yields $-\lambda f(x)\leq f(z)-f(\lambda x+z(1-\lambda ))-\lambda f(z)$ $\Rightarrow -\lambda f(x)-f(z)+\lambda f(z)\leq -f(\lambda x+z(1-\lambda ))$ $\Rightarrow \lambda f(x)+f(z)-\lambda f(z)\geq f(\lambda x+z(1-\lambda ))$ $\Rightarrow \lambda f(x)+f(z)(1-\lambda )\geq f(\lambda x+z(1-\lambda ))$ but since the initial choice of $x,y,z$ was arbitrary (up to ordering) this holds for any $x<z$. We must still address the case corresponding to $x>z$. This is easy though.

For let $x>z$. Then by what we have proven already $\lambda f(z)+f(x)(1-\lambda )\geq f(\lambda z+x(1-\lambda ))$ now let $\alpha =1-\lambda$ so it follows from the above that $(1-\alpha )f(z)+f(x)(1-1+\alpha )\geq f((1-\alpha )z+x(1-1+\alpha ))$ $\Rightarrow (1-\alpha )f(z)+\alpha f(x)\geq f((1-\alpha )z+\alpha x)$

Thus if $f'$ is monotonically increasing $\lambda f(x)+f(z)(1-\lambda )\geq f(\lambda x+z(1-\lambda ))$ for $\forall x,y\in (a,b)$, $\lambda \in (0,1)$

Now we prove ($\Leftarrow$) that $f$ convex implies $f'$ is monotonically increasing.

The proof of this direction follows similarly to the previous. Let's assume $f$ is convex.

Then $\lambda f(x)+f(z)(1-\lambda )\geq f(\lambda x+z(1-\lambda ))$ for $x<z$ (Note: we must have $x<z$ by (*)) and $x,z\in (a,b)$, $\lambda \in (0,1)$. Following the previous reasoning in reverse we see that ${\frac {f(y)-f(x)}{y-x}}\leq {\frac {f(z)-f(y)}{z-y}}$ for $\forall y\in (x,z)$.

So in the limit $y\to x$ we have $f'(x)\leq {\frac {f(z)-f(x)}{z-x}}$ and in the limit $y\to z$ we have ${\frac {f(z)-f(x)}{z-x}}\leq f'(z)$ thus, combining these statements we have $f'(x)\leq {\frac {f(z)-f(x)}{z-x}}\leq f'(z)$ so $f'$ is monotonically increasing as desired.

Now it only remains to prove that (assuming $f''(x)$ exists for $x\in (a,b)$) $f$ is convex if and only if $f''(x)\geq 0$ for $\forall x\in (a,b)$.

$f$ is convex $\Leftrightarrow$ $f'$ is monotonically increasing by the previous proof. Moreover, $f'$ is monotonically increasing $\Leftrightarrow$ $f''\geq 0$ (proved in class).

So we have shown that (assuming $f''(x)$ exists for $x\in (a,b)$) $f$ is convex if and only if $f''(x)\geq 0$ for $\forall x\in (a,b)$ as desired.