Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 8 November 22, 2004

Section 25/16 a) We assume that f is differentiable at c, so according to the definition used in the book, lim_{x→c} \frac{f(x)−f(c)}{x−c} = f′(c) exists. Introduce the new quantity h = x − c. then x = c + h, and the limit as x → c is the same as the limit as h → 0. Hence the above limit becomes

f′(c) = lim_{h→0} \frac{f(c + h) − f(c)}{h}.

b) From part a, we have since f is differentiable at c, f′(c) = lim_{h→0} \frac{f(c+h)−f(c)}{h}. Replacing h by −h in the above, we also have f′(c) = lim_{h→0} \frac{f(c−h)−f(c)}{−h} = lim_{h→0} \frac{f(c)−f(c−h)}{h}. Hence by the limit sum rule,

f′(c) = \frac{1}{2} ( lim_{h→0} \frac{f(c + h) − f(c)}{h} + lim_{h→0} \frac{f(c) − f(c − h)}{h})

= lim_{h→0} \frac{f(c + h) − f(c) + f(c) − f(c − h)}{2h}

= lim_{h→0} \frac{f(c + h) − f(c − h)}{2h},

which is what we wanted to show.

c) Since f is differentiable at c, by part a, f′(c) = lim_{h→0} \frac{f(c+h)−f(c)}{h}. By Theorem 20.8, if we take a sequence h_{n} of h-values converging to 0, with h_{n} ≠ 0 for all n, then f′(c) = lim_{n→∞} \frac{f(c+h_{n})−f(c)}{h_{n}}. So let h_{n} = 1/n. We then have

f′(c) = lim_{n→∞} \frac{f(c + 1/n) − f(c)}{1/n} = lim_{n→∞} n(f(c + 1/n) − f(c)),

which is what we wanted to show.

d) The converse of part b would say: If lim_{h→0} \frac{f(c+h)−f(c−h)}{2h} exists, then f is differentiable and this limit gives the derivative of f at c. To see that this is false, consider an even function that is not differentiable at c = 0, for example, f(x) = |x|. We have

lim_{h→0} \frac{|0 + h| − |0 − h|}{2h} = lim_{h→0} \frac{|h| − |h|}{2h} = 0.

(Geometrically, the secant lines through equally spaced points on either side of zero are always horizontal, so this limit is zero.) But we know from problem 25.7 that f′(0) does not exist.

The converse of part c would say: If lim_{n→∞} n(f(c + 1/n) − f(c)) exists, then f is differentiable and this limit gives the derivative of f at c. To see that this is false, consider a function that is not differentiable at c = 0, but whose difference quotient does have a well-defined one-sided limit as h → 0^{+} (note that 1/n > 0, so this limit basically only takes into account what is happening to the right of c on the graph of f). For example, consider f(x) = |x| again. We have

lim_{h→0+} \frac{|0 + h| − |0|}{h} = lim_{h→0^{+}} 1 = 1.

But we know from problem 25.7 that f′(0) does not exist because the other one-sided limit gives −1.

Section 25/6. First a comment about a proposed solution that a number of people tried to use. If you assume that f is increasing on I and take x_{1} < x_{2}, then it is true that the Mean Value Theorem gives f′(c) = \frac{f(x_{2})−f(x_{1})}{x_{2}−x_{1}} ≥ 0 for some c between x_{1} and x_{2}. Unfortunately, there is no reasonable way, given what we know, to use this to show that f′(x) ≥ 0 for all x ∈ I. The problem is that we have no “control” over which c’s appear in these equations. There’s nothing that says every x ∈ I has to appear in one of them.

So a complete solution for part a has to proceed a different way. First we will show that if f is increasing on I then f′(x) ≥ 0 for all x ∈ I by a direct argument. Take any x ∈ I. Then the definition of f′(x) says f′(x) = lim_{x_{1}→x} \frac{f(x_{1})−f(x)}{x_{1}−x}. Since we assume f is increasing, the difference quotient \frac{f(x_{1})−f(x)}{x_{1}−x} ≥ 0 for all x_{1}. Any limit of a nonnegative function is nonnegative (see Exercise 20.12, for instance). Hence, f′(x) = lim_{x_{1}→x} \frac{f(x_{1})−f(x)}{x_{1}−x} ≥ 0. Next, we show the reverse implication, and this is where the Mean Value Theorem comes in. Assume f′(x) ≥ 0 for all x ∈ I, and let x_{1} < x_{2} in I. Applying the MVT to f on the interval [x_{1}, x_{2}], we get that there exists some c ∈ (x_{1}, x_{2}) such that 0 ≤ f′(c) = \frac{f(x_{2})−f(x_{1})}{x_{2}−x_{1}}. Since x_{2} > x_{1}, this implies f(x_{2}) ≥ f(x_{1}).

Part b is parallel to this and will be omitted here.

Section 26/8. Since f is assumed differentiable on (0, 1), by the quotient rule for derivatives, g(x) = f(x)/x is also differentiable on (0, 1), and

g′(x) = \frac{xf′(x) − f(x)}{x^{2}}

for all x ∈ (0, 1). By Exercise 26.6 above, it suffices to show that g′(x) ≥ 0 for all x ∈ (0, 1). By the MVT, for all x ∈ (0, 1), there exists a c ∈ (0, x) such that f′(c) = \frac{f(x)−f(0)}{x−0} = \frac{f(x)}{x}. Since f′ is assumed to be increasing, since x > c, we have f′(x) ≥ f′(c). Hence f′(x) ≥ \frac{f(x)}{x} for all x ∈ (0, 1). But then

\frac{xf′(x)}{x^{2}} ≥ \frac{f(x)}{x^{2}} ⇒ \frac{xf′(x) − f(x)}{x^{2}} ≥ 0

which shows that g′(x) ≥ 0 for all x ∈ (0, 1).