(a) Define what it means to say that f is a bounded function

Advanced Calculus Handout 2 March 21, 2011 1. Let f be a function with domain D ⊂ R^p and range in R^q.

(a) Define what it means to say that f is a bounded function.

(b) If f is bounded, define the supremum norm of f . Check that (i) kf k_D ≥ 0, and kf k_D = 0 if and only if f (x) = 0 ∀x ∈ D; (ii) kcf k_D = |c|kf k_D; (iii) kf + gk_D ≤ kf k_D+ kgk_D ∀f, g ∈ B_pq(D), which implies that f → kf k_D is a norm on B_pq(D).

2. Let {f_n} be a sequence of functions with domain D ⊂ R^p and range in R^q. (a) Define what it means to say that {fn} is pointwise convergent on D.

(b) Define what it means to say that {f_n} is uniformly convergent on D.

(c) Define what it means to say that {f_n} is Not uniformly convergent on D.

3. (a) For each n ∈ N, let fn be defined for x > 0 by f_n(x) = 1

nx. For what values of f does

n→∞lim fn(x) exist? Is the convergence uniform on the entire set of convergence? Is the convergence uniform for x ≥ 1?

(b) Show that, if we define f_n on R by fn(x) = nx

1 + n²x², then lim

n→∞f_n(x) exists for all x ∈ R.

Is this convergence uniform on R?

(c) Show that lim

n→∞(cos πx)²ⁿ exists for all x ∈ R. Is the limiting function contiuous? Is this convergence uniform?

(d) Let f_n be defined on the interval I = [0, 1] by the formula

f_n(x) =

(1 − nx, if 0 ≤ x ≤ 1/n 0 if 1/n < x ≤ 1.

Show that lim

n→∞f_n(x) exists on I. Is this convergence uniform on I?

(e) Let f_n be defined on the interval I = [0, 1] by the formula

f_n(x) =

(nx, if 0 ≤ x ≤ 1/n

n(1 − x)/(n − 1) if 1/n < x ≤ 1.

Show that lim

n→∞f_n(x) exists on I. Is this convergence uniform on I? Is the convergence uniform on [c, 1] for c > 0?

4. Let f be a function with domain D ⊂ R^p and range in R^q. (a) Define what it means to say that f is a Lipschitz function.

(b) Define what it means to say that f is a continuous function on D.

(c) Is a Lipschitz function always continuous?

(d) Is a linear function f : R^p → R^q always Lipschitz? [May write f (x) = Ax + b, , where A is a q × p constant matrix, and b is a q × 1 vector]

(e) Is a Lipschitz function always differentiable?

(f) Let f be a differentiable function defined on an open subset U of R^p. Is f Lipschtz on U ? Is f Lipschitz on any compact subset K of U ?

5. Let X = {x_mn} be a double sequence in R^p.

(a) Define what it means to say that x is a limit of {x_mn}.

Advanced Calculus Handout 2 (Continued) March 21, 2011 (b) For each m ∈ N, let Ym = {x_mn} be a sequence in R^p which converges to y_m. Define what

it means to say that {Y_m | m ∈ N} are uniformly convergent.

6. Let f : [a, b] → R be differentiable with 0 < m ≤ f⁰(x) ≤ M for x ∈ [a, b] and let f (a) < 0 < f (b).

Given x₁ ∈ [a, b], define the sequence {x_n} by x_n+1 = x_n− 1

Mf (x_n), for n ∈ N. Prove that this sequence is well-defined and converges to the unique root ¯x of the equation f (¯x) = 0 in [a, b]. [Hint: Consider φ : [a, b] → R defined by φ(x) = x − f (x)

M . Does φ map [a, b] into [a, b] = domain(f ) ?]

Solution: For any x ∈ [a, b], since 0 < f⁰(t) ≤ M for all t ∈ [a, b], we have f (x) − f (a) = Z x

a

f⁰(t)dt ≤ M Z x

a

dt = M (x − a) which implies that a = x − f (x)

M + f (a)

M ≤ x − f (x)

M , where we have used the assumption that f (a) < 0 in the last inequality.

Similarly, f (b) − f (x) = Z b

x

f⁰(t)dt ≤ M Z b

x

dt = M (b − x) which implies that b = f (b)

M + x − f (x)

M + ≥ x − f (x)

M , where we have used the assumption that f (b) > 0 in the last inequality.

Hence, the function φ(x) = x−f (x)

M maps [a, b] into [a, b], and the sequence x_n+1 = x_n− 1

Mf (x_n), for n ∈ N, is well defined if x1 ∈ [a, b].

For any x, y ∈ [a, b], since |φ(x) − φ(y)| = |x − y −f⁰(c)

M (x − y)|, for some c lying between x, and y.

Hence, |φ(x) − φ(y)| ≤ (1 −m

M)|x − y| and φ is Lipschitz with Lipschitz constant 0 ≤ 1 − m M < 1.

Hence, φ is a contraction and has a unique fixed point ¯x ∈ [a, b].

Since |xn+1− ¯x| = |φ(xn) − φ(¯x)| ≤ (1 − m

M)|xn− ¯x| ≤ (1 − m

M)ⁿ|x1− ¯x| ≤ (1 − m

M)ⁿ(b − a),

¯

x = lim

n→∞x_n+1= lim

n→∞x_n− lim

n→∞

f (x_n)

M = ¯x − f (¯x)

M . Hence, f (¯x) = 0.

7. For each integer k ≥ 1, let f_k: R → R be differentiable satisfying |fk⁰(x)| ≤ 1 for all x ∈ R, and f_k(0) = 0.

(a) For each x ∈ R, prove that the set {fk(x)} is bounded.

Solution: For each x ∈ R, since |fk(x)| = |f_k(x) − f_k(0)| ≤ |f_k⁰(c_k) (x − 0)| ≤ |x|, where c_k lies between x and 0, the set {f_k(x)}^∞_k=1 is bounded.

(b) Use Cantor’s diagonal method to show that there is an increasing sequence n₁ < n₂ < n₃ <

· · · of positive integers such that, for every x ∈ Q, we have {fnk(x)} is a convergent sequence of real numbers.

Solution Let Q = {x1, x₂, · · · }. Since {f_k(x₁)} is bounded, we can extract a convergent subsequence, denoted {f_k¹(x₁)}, out of {f_k(x₁)}. Next, the boundedness of {f_k¹(x₂)} implies that we can extract a convergent subsequence, denoted by {f_k²(x₂)}, out of {f_k¹(x₂)}. Con- tinuing this way, the boundedness of {f_k^j(x_j+1)} implies that we can extract a convergent subsequence {f_k^j+1(x_j+1)}, out of {f_k^j(x_j+1)} for each j ≥ 1. Let f_nk = f_k^k for each k ≥ 1.

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

8. Let F be a bounded and uniformly equicontinuous collection of functions on D ⊂ R^p to R and let f^∗ be defined on D → R by

f^∗(x) = sup{f (x) | f ∈F , x ∈ D}

Show that f^∗ is continuous on D to R. Show that the conclusion may fail if F is not uniformly equicontinuous.

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