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Theorem 1 (The Triangle Inequality). For all x, y ∈ R we have
|x + y| ≤ |x| + |y|.
Proposition 2 (The Archimedean property). For each x ∈ R there exists an n ∈ N such that n > x.
Definition 3. A Cauchy sequence is a sequence of real numbers (an) such that for every ε > 0 there exists a positive integer N such that
|am − an| < ε whenever m, n ≥ N.
Proposition 4. For each n ≥ 1 let Hn := 1 + 12 + 13 + · · · + (^1) n. Then the sequence (H 1 , H 2 , H 3 ,.. .) is not a Cauchy sequence.
Proof. For each n ∈ N, consider the difference
|H 2 n − Hn| = 1 n + 1 +^
1 n + 2 +^
1 n + 3 +^ · · ·^ +^
1 2 n.
Notice that there are exactly n terms in the sum, and each term is larger than (or equal to) the smallest one (^21) n. Thus
|H 2 n − Hn| ≥ 1 2 n +^ · · ·^ +^
1 2 n =^
1
So for each N ≥ 1 let m = 2N and n = N ; then |Hm − Hn| ≥ 12.
Definition 5. A sequence (an) is bounded if there exists M such that |an| ≤ M for all n ≥ 1.
Proposition 6. Every Cauchy sequence is bounded.
Proof. Suppose that (an) is a Cauchy sequence. Taking ε = 1, this means that there is some N ∈ N for which |am − an| ≤ 1 for all m, n ≥ N. This splits our sequence into a finite piece (indices 1 through N ) and an infinite piece (index N and after). By Lemma ??, the finite piece is bounded, say by M. Using the Triangle Inequality we find that
|an| = |an − aN + aN | ≤ |an − aN | + |aN | < 1 + M.
It follows that (an) is bounded by M + 1.
Definition 7. Two sequences (an) and (bn) are equivalent if for each ε > 0 there exists a natural number N (which can depend on ε) such that
|an − bn| < ε whenever n ≥ N.
Proposition 8 (Q is dense in R). For every pair of real numbers x, y with x < y there exists a rational number q such that x < q < y.
Definition 9. Let (an) be a sequence of real numbers. We say that (an) converges to L ∈ R (and we write lim n→∞ an = L) if for every ε > 0 there exists a natural number N for which
|an − L| < ε for all n ≥ N. 1
Theorem 10 (R is complete). Every Cauchy sequence in R converges to a real number.
Definition 11. Suppose that E ⊆ R is a set of real numbers and let M, m ∈ R. We say that M is a least upper bound for E if (a) M is an upper bound for E and (b) if M ′^ is another upper bound for E, then M ′^ ≥ M. Similarly, we say that m is a greatest lower bound for E if (a) m is a lower bound for E and (b) if m′^ is another lower bound for E then m′^ ≤ m.
Theorem 12 (Least upper bound property). Let E be a non-empty subset of R. If E has an upper bound, then it has a least upper bound.
Proof. Since E is non-empty, there is some element a 1 ∈ E, and since E has an upper bound, we can choose some upper bound b 1. Clearly a 1 ≤ b 1. We will define two sequences (ai) and (bi) inductively such that an ≤ bn for all n. For each n ≥ 1, suppose that we have constructed an and bn such that an ≤ bn. Let K = an+ 2 bn be the average of an and bn. Note that an ≤ K ≤ bn. If K is an upper bound
for E, let an+1 = an and let bn+1 = K = an+ 2 bn. Then |bn+1 − an+1| = |an+ 2 bn− an| = bn− 2 an. If K is not an upper bound for E, then there is some element x ∈ E such that x > K (otherwise K would be an upper bound). In that case, let an+1 = x and let bn+1 = bn. Then |bn+1 − an+1| = bn − x < bn − K = bn− 2 an. Note that in either case |bn+1 − an+1| ≤ bn− 2 an. Also in either case an is increasing and bn is decreasing. Since an ∈ E for all n and bn is an upper bound for E for all n, we have a 1 ≤ a 2 ≤ a 3 ≤ · · · ≤ b 3 ≤ b 2 ≤ b 1.
Using the inequality |bn+1 − an+1| ≤ bn− 2 anone can show by induction that
|bn+1 − an+1| ≤ b 1 − a 1 2 n^.^ (1)
We will show that the sequences (an) and (bn) are Cauchy. Let ε > 0 be given. Choose N ≥ 1 such that 2N^ > b^1 − ε a^1. Let m, n ≥ N. Without loss of generality, assume that m ≥ n. Then am ≥ an, and we have
|am − an| = am − an ≤ bn − an ≤
b 1 − a 1 2 n^
< ε.
Thus (an) is Cauchy. A very similar argument shows that (bn) is Cauchy. Using (1) one can show that (an) and (bn) are equivalent sequences. Therefore they have the same limit, say L. I claim that L is the least upper bound of E. On the one hand, if x ∈ E then x ≤ bn for all n since every bn is an upper bound for E. Thus x ≤ L. Since x was an arbitrary element of E, we conclude that L is an upper bound for E. On the other hand, if L′^ is any other upper bound, then L′^ ≥ an for all n since every an is an element of E. Thus L′^ ≥ L. It follows that L is the least upper bound of E.
Proposition 13 (Bounded monotonic sequences are convergent). If a sequence is increasing and bounded above then it converges.
Proof. Suppose that (an) is increasing and bounded above. Let E = {an : n ≥ 1 }. Then (an) has a least upper bound which we will call s = sup an. We will show that lim an = s. Let ε > 0 be given. Since s is the least upper bound, s − ε is not an upper bound for E. So there exists some N ≥ 1 such that aN > s − ε. Since (an) is increasing, it follows that an ≥ aN > s − ε for all n ≥ N. Thus, for all n ≥ N we have
|an − s| = s − an ≤ s − aN < s − (s − ε) = ε.
Proposition 21 (Alternating series test). If (an) is a decreasing sequence of nonnegative real numbers and lim an = 0 then the alternating series
(−1)n+1an converges.
Proof. Let (sn) denote the sequence of partial sums. Let (en) be the subsequence of even- indexed terms and (on) the subseqeunce of odd-indexed terms. Then on = s 2 n− 1 and en = s 2 n:
(sn) = o 1 , e 1 , o 2 , e 2 , o 3 , e 3 ,....
The even-indexed terms are increasing since
en+1 − en = s 2 n+2 − s 2 n = −a 2 n+2 + a 2 n+1 ≥ 0.
Similarly, the odd-indexed terms are decreasing since
on+1 − on = s 2 n+1 − s 2 n− 1 = a 2 n+1 − a 2 n ≤ 0.
I claim that every even-indexed term is less than or equal to every odd-indexed term:
em ≤ on for all m, n ∈ N.
First, en ≤ on+1 ≤ on because on+1 − en = s 2 n+1 − s 2 n = a 2 n+1 ≥ 0 and (on) is decreasing. Now, if m ≤ n, then because (en) is increasing, em ≤ en ≤ on. On the other hand, if m ≥ n then on ≥ om ≥ em. To summarize,
e 1 ≤ e 2 ≤ e 3 ≤ · · · ≤ o 3 ≤ o 2 ≤ o 1.
Since (en) is increasing and bounded above, it converges, say to Le. Similarly, (on) converges, say to Lo. But
Lo − Le = lim n→∞ on+1 − lim n→∞ en = lim n→∞ (on+1 − en) = lim n→∞ a 2 n+1 = 0.
Thus Lo = Le. It follows that (sn) converges.
Definition 22. Let X ⊆ R. A function f : X → R is continuous at the point a ∈ X if, for every sequence (an) (with an ∈ X) converging to a, the sequence (f (an)) converges to f (a). Put another way, f is continuous at a if
lim n→∞ f (an) = f
lim n→∞ an
for every sequence (an) that converges to a.
Definition 23. Let X ⊆ R. A function f : X → R is continuous at the point a ∈ X if, for every ε > 0 there exists a δ > 0 such that
|f (x) − f (a)| < ε whenever x ∈ X and |x − a| < δ.
Theorem 24 (Extreme Value Theorem). Let f : [a, b] → R be a continuous function. Then f is bounded (i.e. there exists a real number M such that |f (x)| ≤ M for all x ∈ [a, b]). Furthermore, f attains its maximum and minimum values on [a, b] (i.e. there exist x 0 , y 0 ∈ [a, b] such that f (x 0 ) ≤ f (x) ≤ f (y 0 ) for all x ∈ [a, b]).
Proof. We argue by contradiction. Suppose that f is not bounded on [a, b]. Then for each n ∈ N there exists a number xn ∈ [a, b] such that |f (xn)| > n. So we get a sequence (xn) which is bounded (it’s always inside [a, b]). By the Bolzano-Weierstrass theorem, there exists a convergent subsequence (xnk ) of (xn), say that lim xnk = L. Then L ∈ [a, b], so by assumption f is continuous at L. So we have a sequence (xnk ) which converges to L but the sequence (f (xnk )) doesn’t converge to f (L) (since |f (xnk )| → +∞). This contradicts the continuity of f , so we must have that f is bounded.
Now let M := sup{f (x) : x ∈ [a, b]}. We just proved that M is a finite number. For each n ∈ N there exists a number yn ∈ [a, b] such that M − (^1) n ≤ f (yn) ≤ M (otherwise M wouldn’t be the sup). By the squeeze theorem, lim f (yn) = M. We aren’t guaranteed that (yn) is a convergent sequence, but it is bounded (always in [a, b]) so Bolzano-Weierstrass gives us a convergent subsequence (ynk ). Say lim ynk = y 0. Since f is continuous at y 0 we have f (y 0 ) = f (lim ynk ) = lim f (ynk ) = lim f (yn) = M.
(In the third equality we used that if a sequence is convergent, then every subsequence converges to the same limit.) So f attains its maximum M. The assertion for the minimum is similar.
Theorem 25 (Intermediate Value Theorem). Let I ⊆ R be an interval and suppose that f : I → R is continuous. Let a, b ∈ I such that a < b. Then for every y between f (a) and f (b) (i.e. either f (a) < y < f (b) or f (a) > y > f (b)) there exists at least one x ∈ (a, b) such that f (x) = y.
Proof. Assume that f (a) < y < f (b) (the other case is similar). We will construct two sequences (an) and (bn) which converge to x 0 such that f (an) ≥ y and f (bn) ≤ y. Let S = {x ∈ [a, b] : f (x) < y}. Then S is not empty because a ∈ S. Since S is also bounded above (by b) it has a least upper bound, say x 0 = sup S. For each n ∈ N, the number x 0 − (^) n^1 is not an upper bound for S so there exists bn ∈ S such that x 0 − (^1) n < bn ≤ x 0. By the squeeze theorem, lim bn = x 0. Since f is continuous and since f (bn) < y we have
f (x 0 ) = f (lim bn) = lim f (bn) ≤ y.
Now for each n ∈ N let an = min{b, x 0 + (^) n^1 }. Then for each n, an ∈/ S so f (an) ≥ y. Again by the squeeze theorem, lim an = x 0 so we have
f (x 0 ) = f (lim an) = lim f (an) ≥ y.
Thus f (x 0 ) = y.
Definition 26. A function f : X → R is uniformly continuous on X if for every ε > 0 there exists a δ > 0 such that
|f (x) − f (y)| < ε whenever x, y ∈ X and |x − y| < δ.
Proposition 27. If f : X → R is uniformly continuous on X and if (an) is a Cauchy sequence then (f (an)) is a Cauchy sequence.
Proof. Let (an) be a Cauchy sequence and let ε > 0. Since f is uniformly continuous on X, there exists a δ > 0 such that |f (x) − f (y)| < ε
whenever x, y ∈ X and |x − y| < δ. Since (an) is Cauchy, there exists and N ≥ 1 such that
|am − an| < δ
for all m, n ≥ N. Thus, for such m, n we have
|f (am) − f (an)| < ε
so (f (an)) is a Cauchy sequence.
Proposition 28. A function f : (a, b) → R can be extended to a continuous function on [a, b] if and only if it is uniformly continuous on (a, b).
Proposition 33 (Product rule). Let f and g be functions that are differentiable at a. Then f g is differentiable at a and
(f g)′(a) = f (a)g′(a) + f ′(a)g(a).
Proof. Adding zero in a clever way, we compute
lim x→a
f (x)g(x) − f (a)g(a) x − a
= lim x→a
f (x)g(x) − f (x)g(a) + f (x)g(a) − f (a)g(a) x − a
= lim x→a f (x)
g(x) − g(a) x − a
f (x) − f (a) x − a = f (a)g′(a) + f ′(a)g(a),
as desired.
Theorem 34 (Rolle’s Theorem). Suppose that f is continuous on [a, b] and differentiable on (a, b) and that f (a) = f (b). Then there exists a point c ∈ (a, b) such that f ′(c) = 0.
Proof. By the Extreme Value Theorem, since f is continuous on the closed interval [a, b] it attains its maximum and minimum on [a, b]. So there exist y, z ∈ [a, b] such that f (y) ≤ f (x) ≤ f (z) for all x ∈ [a, b]. Suppose that z is an interior point, i.e. z ∈ (a, b). Let (zn) be a sequence in (a, z) ∪ (z, b) converging to z and let (z n+ ) and (z n− ) be the subsequences of (zn) such that z n+ > z and z− n < z. Then (since f (z) is a maximum)
lim n→∞
f (z n+ ) − f (z) z n+ − z
≤ 0 and lim n→∞
f (z− n ) − f (z) z n− − z
(Both limits exist because f is differentiable at z.) Since subsequences of a convergent sequence converge to the same limit, we see that
lim n→∞
f (zn) − f (z) zn − z
≤ 0 and lim n→∞
f (zn) − f (z) zn − z
It follows that f ′(z) = 0. By the same reasoning, we have f ′(y) = 0 if y ∈ (a, b). The only case left is if y and z are both endpoints of [a, b]. But since f (a) = f (b) this forces f (y) = f (x) = f (z) for all x ∈ [a, b], so f ′(x) = 0 for all x ∈ (a, b).
Theorem 35 (The Mean Value Theorem). Suppose that f is continuous on [a, b] and dif- ferentiable on (a, b). Then there exists a c ∈ (a, b) such that
f ′(c) =
f (b) − f (a) b − a
Proof. Define
g(x) = f (x) −
f (b) − f (a) b − a
(x − b) + f (b)
The function in parentheses is just the equation of the secant line connecting (a, f (a)) and (b, f (b)). We compute g(a) = 0 and g(b) = 0.
Clearly g is continuous on [a, b] and differentiable on (a, b), so Rolle’s theorem applies. So there is some point c ∈ (a, b) such that g′(c) = 0. On the other hand,
g′(c) = f ′(c) −
f (b) − f (a) b − a
This proves the theorem.
Definition 36. Suppose that f is bounded on [a, b]. A partition P of [a, b] is a finite ordered set P = {a = x 0 < x 1 < x 2 <... < xn = b}.
The intervals [xk− 1 , xk] are called subintervals of P. For each subinterval, let
mk = inf{f (x) : x ∈ [xk− 1 , xk]}, Mk = sup{f (x) : x ∈ [xk− 1 , xk]}.
The lower Darboux sum L(f, P ) of f with respect to P is
L(f, P ) =
∑^ n
k=
mk(xk − xk− 1 )
and the upper Darboux sum U (f, P ) of f with respect to P is
U (f, P ) =
∑^ n
k=
Mk(xk − xk− 1 ).
We define L(f ) = sup{L(f, P ) : P is a partition of [a, b]}
and U (f ) = inf{U (f, P ) : P is a partition of [a, b]},
and we say that f is integrable on [a, b] if L(f ) = U (f ). In that case we write ∫ (^) b
a
f = L(f ) = U (f ).
Proposition 37 (Cauchy criterion for integrals). Suppose that f is bounded on [a, b]. Then f is integrable if and only if for every ε > 0 there exists a partition P of [a, b] such that
U (f, P ) − L(f, P ) < ε.
Proof. (⇐) Suppose that such a partition P exists for each ε > 0. Then
U (f ) − L(f ) ≤ U (f, P ) − L(f, P ) < ε.
Since ε is arbitrary, this implies that U (f ) = L(f ). (⇒) Now suppose that f is integrable. Let ε > 0 be given. Since U (f ) is the greatest lower bound of the U (f, P ) over all partitions P , the number U (f ) + ε 2 is no longer a lower bound, so there exists a partition P 1 such that
U (f, P 1 ) < U (f ) +
ε
Similarly, there exists a partition P 2 such that
L(f, P 2 ) > L(f ) − ε
Let P = P 1 ∪ P 2. Since integrability means that U (f ) = L(f ), we have
U (f, P ) − L(f, P ) ≤ U (f, P 1 ) − L(f, P 2 ) = U (f, P 1 ) − U (f ) + L(f ) − L(f, P 2 ) < ε 2 +^
ε 2 =^ ε. This completes the proof.
Proof. Let M be such that |g(x)| ≤ M for all x ∈ [a, b]. For x, y ∈ [a, b] we have
|F (y) − F (x)| =
∫ (^) y
a
f (t) dt −
∫ (^) x
a
f (t) dt
∫ (^) y
x
f (t) dt
∫ (^) y
x
|f (t)| dt ≤ M |x − y|.
Therefore F is Lipschitz, so it is uniformly continuous on [a, b]. Suppose that f is continuous at c ∈ (a, b). Observe that
f (c) =
1 x − c
∫ (^) x
c
f (c) dt
and F (x) − F (c) x − c
1 x − c
∫ (^) x
c
f (t) dt,
so F (x) − F (c) x − c
− f (c) = 1 x − c
∫ (^) x
c
(f (t) − f (c)) dt.
Let ε > 0 be given. Since f is continuous at c, there exists a δ > 0 such that |f (t)−f (c)| < ε whenever |t − c| < δ. For the integral above we only care about values of t between c and x, and if |x − c| < δ then certainly |t − c| < δ. For such x it follows that ∣ ∣ ∣ ∣
F (x) − F (c) x − c
− f (c)
∣ ≤^
1 |x − c|
∫ (^) x
c
|f (t) − f (c)| dt < 1 |x − c|
∫ (^) x
c
ε dt = ε.
Therefore F ′(c) = f (c).
