Chapter 5Edit

6Edit

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

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

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

11Edit

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

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

14Edit

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

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

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

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

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

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

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

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

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

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

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

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