# Chapter 5Edit

## 6Edit

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.

## 11Edit

Since the denominator of $\lim_{h \to 0}\frac{f(x+h)+f(x-h)-2 f(x)}{h^2}$ goes to zero as $h \to 0$ we can employ L'Hopital's rule. Using this we get $\lim_{h \to 0}\frac{f(x+h)+f(x-h)-2 f(x)}{h^2} = \lim_{h \to 0} \frac{f'(x+h)-f'(x-h)}{2 h}$ 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 $\lim_{h \to 0}\frac{f(x+h)+f(x-h)-2 f(x)}{h^2}=\lim_{h \to 0}\frac{h^2}{h^2}=1$

## 14Edit

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 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. 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 (Note: we must have $x 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.