(a) (8 points) Prove that f is differentiable at a point x ∈ [a, b] if and only if there exists a number l∈ R such that t→x

1. Let f : [a, b] → R be a real-valued function defined on [a, b] ⊂ R.

(a) (8 points) Prove that f is differentiable at a point x ∈ [a, b] if and only if there exists a number l∈ R such that

t→x ; t∈[a,b]lim

| f (t) − f (x) − l(t − x)|

|t − x| = 0.

Solution: If f is differentiable at a point x ∈ [a, b], then f⁰(x) = lim

t→x ; t∈[a,b]

f(t) − f (x)

t− x exists . Thus, we have

t→x ; t∈[a,b]lim

f(t) − f (x) − f⁰(x)(t − x)

t− x = 0 ⇐⇒ lim

t→x ; t∈[a,b]

| f (t) − f (x) − f⁰(x)(t − x)|

|t − x| = 0.

If there exists a number l ∈ R such that lim

t→x ; t∈[a,b]

| f (t) − f (x) − l(t − x)|

|t − x| = 0,

then we have

t→x ; t∈[a,b]lim

f(t) − f (x) − l(t − x)

t− x = 0 ⇐⇒ lim

t→x ; t∈[a,b]

f(t) − f (x)

t− x = l ∈ R, i.e. f is differentiable at a point x ∈ [a, b] with f⁰(x) = l.

(b) (8 points) Show that if f is differentiable at a point x ∈ [a, b] then f is continuous at x.

Solution: If f is differentiable at a point x ∈ [a, b], then f⁰(x) = lim

t→x ; t∈[a,b]

f(t) − f (x)

t− x exists and

lim

t→x ; t∈[a,b] f(t) − f (x)

= lim

t→x ; t∈[a,b]

f(t) − f (x)

t− x
t− x

= lim

t→x ; t∈[a,b]

f(t) − f (x)

t− x lim

t→x ; t∈[a,b] t− x

= f⁰(x) · 0

= 0.

Hence, lim

t→x ; t∈[a,b]f(t) = f (x) and f is continuous at x.

2. (a) (8 points) Let f be defined for all real x, and suppose that

| f (x) − f (y)| ≤ (x − y)² for all x, y ∈ R.

Prove that f is constant.

Solution: Suppose that

| f (x) − f (y)| ≤ (x − y)² = |x − y|² for all x, y ∈ R.

Then

x→ylim

| f (x) − f (y)|

|x − y| ≤ lim

x→y|x − y| = 0 ∀ y ∈ R and

f⁰(y) = lim

x→y

f(x) − f (y)

x− y = 0 ∀ y ∈ R.

This proves that f is constant.

(b) (8 points) Suppose g is a real function on R, with bounded derivative (say |g⁰| ≤ M). For ε > 0, and define

f(x) = x + εg(x).

Prove that f is one-to-one if ε is small enough.

Solution: By choosing 0 < ε < 1

M, since

| f⁰(x)| = |1 + εg⁰(x)| ≥ 1 − ε|g⁰(x)| > 1 − 1

M· M = 0 ∀ x ∈ R =⇒ | f⁰(x)| > 0 ∀ x ∈ R, f is an one-to-one function on R by the Mean Value Theorem.

3. Let E be a subset of a metric space X and let fn: E → C be a sequence of complex-valued functions deined on E.

(a) (8 points) Prove that f_nconverges uniformly to f : E → C on E if and only if

∀ ε > 0, ∃ N = N(ε) ∈ N such that if m, n ≥ N then k fm− f_nk ≤ ε.

Solution:

Suppose that f_nconverges uniformly on E to f

=⇒ ∀ ε > 0, ∃N = N(ε) ∈ N such that if n ≥ N then | fn(x) − f (x)| ≤ε

2∀ x ∈ E

=⇒ ∀ ε > 0, ∃N = N(ε) ∈ N such that if n ≥ N then k fn− f k ≤ε 2

=⇒ ∀ ε > 0, ∃ N = N(ε) ∈ N such that if m, n ≥ N then k f_m− f_nk = k f_m− f + f − f_nk ≤ k f_m− f k + k f − f_nk ≤ε

2+ε 2 = ε.

Suppose that ∀ ε > 0, ∃ N = N(ε) ∈ N such that if m, n ≥ N then k fm− f_nk ≤ ε

=⇒ | f_n(x) − f (x)| ≤ ε ∀ m, n ≥ N, ∀ x ∈ E

=⇒ { f_n(x)} is a Cauchy sequence ∀ x ∈ E

=⇒ Since C is complete, f (x) = lim

n→∞f_n(x) exists ∀ x ∈ E

=⇒ ∀ m ≥ N, ∀ x ∈ E then | f_n(x) − f (x)| ≤ ε

=⇒ f_nconverges uniformly on E to f .

(b) (8 points) Suppose that f_n: E → C is continuous for each n ∈ N and suppose that fn converges uniformly to f : E → C on E. Show that f is continuous.

Solution: For each ε > 0, since fnuniformly to f on E, there exists N = N(ε) ∈ N such that if n ≥ N then | f_n(x) − f (x)| ≤ ε

3 ∀ x ∈ E.

Also, for each x ∈ E since fN: E → C is continuous on E, there exists δ = δ (x, ε) > 0 such that if y ∈ E and d(x, y) < δ then | f_N(x) − f_N(y)| <ε

3 ∀ x ∈ E.

This imples that

if y ∈ E and d(x, y) < δ then

| f (x) − f (y)| = | f (x) − f_N(x) + f_N(x) − f_N(y) + f_N(y) − f (y)|

≤ | f (x) − f_N(x)| + | f_N(x) − f_N(y)| + | f_N(y) − f (y)|

< ε 3+ε

3+ε

= ε. 3

4. Let E be a subset of a metric space X and let f_n: E → C be a sequence of complex-valued functions deined on E.

(a) (4 points) State the definition when we say that f_nis pointwise bounded on E.

Solution: { f_n} is pointwise bounded on E if { f_n(x) | n ∈ N} is bounded for every x ∈ E, i.e.

there exists a function φ : E → R such that | fn(x)| ≤ φ (x) for all x ∈ E and for all n ∈ N.

(b) (4 points) State the definition when we say that f_nis bounded (or uniformly bounded) on E.

Solution: { f_n} is bounded (or uniformly bounded) on E if there exists a constant M such that k f k = sup{ f (x) | x ∈ E} ≤ M, for all n ∈ N.

(c) (4 points) State the definition when we say that f_nis uniformly equicontinuous on E.

Solution: { f_n} is said to be uniformly equicontinuous on E if, for each real number ε > 0 there is a number δ = δ (ε) > 0 such that if x, y belong to E and kx − yk < δ then k f_n(x) − f_n(y)k < ε for all n ∈ N.

5. For each n ∈ N, let fn: R → R be differentiable such that | f_n⁰(x)| ≤ 1 for all x ∈ R and fn(0) = 0.

(a) (8 points) Prove that the set { fn} is is uniformly equicontinuous on R.

Solution: Given ε > 0, since fn: R → R is differentiable and | fn⁰(x)| ≤ 1 for all x ∈ R and for each n ∈ N, choose δ = ε such that if x, y ∈ R and kx − yk < δ , by the Mean Value Theorem, there exists z_n, lying between x and y such that

| f_n(x) − f_n(y)| = | f_n⁰(z_n)(x − y)| ≤ |x − y| < δ = ε.

This imples that { f_n} is is uniformly equicontinuous on R.

(b) (8 points) Prove that the set { f_n} is pointwise bounded on R, i.e. for each x ∈ R, prove that the set { f_n(x)} is bounded.

Solution: For each x ∈ R, since fn: R → R is differentiable on R, fn(0) = 0 and by the Mean Value Theorem, there exists z_n, lying between x and 0 such that

| f_n(x)| = | f_n(x) − f_n(0)| = | f_n⁰(z_n)(x − 0)| ≤ |x − 0| = |x|.

Hence, the set { fn} is pointwise bounded on R.

(c) (8 points) Use the Cantor’s diagonal process to find a subsequence { f_nk} of { f_n} such that { f_nk} is pointwise convergent on Q.

Solution: Let Q = {x1, x₂, · · · }.

Since { f_n(x₁)} is a bounded subset of R, there is a convergent subsequence { fn¹(x₁)} of { f_n(x₁)}.

Next, the boundedness of { f_n¹(x₂)} implies that there is a convergent subsequence { f_n²(x₂)} of { f_n¹(x₂)}.

By continuing the process, the boundedness of { f_n^j(xj+1)} implies that there is a convergent subsequence { f_n^j+1(x_j+1)} of { f_n^j(x_j+1)} for each j ≥ 1.

Let f_nk = f_k^kfor each k ≥ 1.

Then { f_nk} is a subsequence of { f_n} and { f_nk} converges at each x_j∈ Q.

6. For each n ∈ N, let fn: R → R be defined by fn(x) = x 1 + nx².

(a) (8 points) Show that f_nconverges uniformly to f (x) ≡ 0 for all x ∈ R.

Hint: You may use the fact that if a > 0, b > 0 then a + b ≥ 2√

a b ⇐⇒ 1

a+ b ≤ 1 2√

a b to estimate 1

1 + nx².

Solution: For each n ∈ N, since fn(0) = 0 and for each x 6= 0 1 + nx²≥ 2√

n|x| =⇒ 1

1 + nx² ≤ 1 2√

n|x|. Thus, we get

| f_n(x)| ≤ |x|

2√

n|x| ≤ 1 2√

n ∀ x ∈ R.

This imples that for any ε > 0, let N ∈ N be chosen such that 1 2√

N < ε, we have if n ≥ N then

| f_n(x) − 0| = | f_n(x)| ≤ 1 2√

n ≤ 1

2√

N < ε ∀ x ∈ R.

This proves that f_nconverges uniformly to f (x) ≡ 0 for all x ∈ R.

(b) (8 points) Show that the equation

f⁰(x) = lim

n→∞f_n⁰(x) is correct if x 6= 0, but false if x = 0.

Solution: Since f⁰(x) = 0 for all x ∈ R and

f_n⁰(x) = 1

1 + nx²− 2nx²

1 + nx²
2 = 1 − nx² 1 + nx²
2

which implies that

n→∞lim f_n⁰(x) =

(0 = f⁰(x) if x 6= 0 1 6= f⁰(0) if x = 0.