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

Chapter 5Edit


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.


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


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.

Last modified on 19 December 2013, at 01:06