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 $$\lim_{h \to 0}\frac{f(x+h)+f(x-h)-2 f(x)}{h^2}$$ 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 $$\lim_{h \to 0}\frac{f(x+h)+f(x-h)-2 f(x)}{h^2}=\lim_{h \to 0}\frac{h^2}{h^2}=1$$

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 $$x 0$$ 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 $$xz$$. 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.