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Math 508 Exam 2 Jerry L. Kazdan
December 4, 2008 10:30 – 11:
Directions This exam has two parts, Part A has 10 True-False problems (30 points, 3 points each). Part B has 5 traditional problems (70 points, 14 points each). Closed book, no calculators or computers– but you may use one 3′′^ × 5 ′′^ card with notes on both sides.
Part A: True/False (answer only, no reasons). 10 problems, 3 points each).
Circle T or F in in each problem.
1. T F A bounded sequence {an} of real numbers always has a convergent subsequence.
True Rudin, p. 51 Theorem 3.6b)
2. T F A series
n=1 an^ of complex numbers converges if and only if the corresponding sequence of partial sums is bounded. False Example:
(−1)n^.
3. T F A closed and bounded subset of a complete metric space must be compact.
False Example: The unit sphere in ` 2.
4. T F If A and B are compact subsets of a metric space, then A ∪ B is also compact.
True Rudin, p.38 2.35 Corollary
5. T F If M is any metric space and f : M → R is any continuous real-valued function,
then the function g : M → R defined by g(x) := f (x)^2 is always continuous. True Given x 0 ∈ M and > 0, pick δ > 0 so that if d(x, x 0 ) < δ , then |f (x) − f (x 0 )| < . Let M = |f (x 0 )|. Then
|g(x) − g(x 0 )| = |f (x)^2 − f (x 0 )^2 | = |
f (x) − f (x 0 )
f (x) − f (x 0 ) + 2f (x 0 )
| ≤ ( + 2M ).
6. T F If f : X → Y is a continuous map between metric spaces, and f (X) is compact,
then X is compact. False Example: X = R, f (x) = 0 for all x ∈ R.
7. T F A compact subset of a metric space is always complete.
True Rudin, p. 54 (after Definition 3.12)
8. T F Let {xn} be a sequence of points in a metric space. If two subsequences of this
sequence converge, then they must converge to the same number. False Example: In R, {(−1)n}.
9. T F If f : [0, 1] → R is a continuous function and
0 f^ (x)^ dx^ = 0, then^ f^ (x) is positive somewhere and negative somewhere in this interval (unless it is identically zero). True If f is not identically zero, then it is either positive somewhere or negative somewhere (or both). Say it is positive at x 0. Then by continuity, it is positive in a neignborhood of x 0. If f (x) ≥ 0, everywhere, then
0 f^ (x)^ dx >^ 0, a contradiction. Thus^ f^ must be negative at some x 1 – and hence also in a neighborhood of x 1.
10. T F f (x) :=
∑^ ∞
1
sin(3nπx) 2 n^
is a continuous function on R.
True Since
∣∣ sin(3nπx) 2 n
2 n^
, by the Weierstrass M-Test the series converges uniformly and absolutely – and hence to a continuous function.
Part B: Traditional Problems (5 problems, 14 points each)
B–1. Let f : [− 2 , 2] be a smooth function with the property that
f (−1) = 1, f (0) = 0, f (1) = 2.
Show that at some point c ∈ (− 1 , 1) we have f ′′(c) > 0. In fact, find an explicit constant m > 0 so that f ′′(c) ≥ m.
Solution By the mean value theorem applied to the intervals [− 1 , 0] and [0, 1] there are points a ∈ (− 1 , 0) and b ∈ (0, 1) so that f ′(a) = −1 and f ′(b) = 2. Applying the mean value theorem to f ′^ , we conclude there is a point c ∈ (a, b) such that f ′′(c) = 3/(b − a) > 3 /2.
B–2. Let A(t) and B(t) be n × n matrices that are differentiable for t ∈ [a, b] and let t 0 ∈ (a, b). Directly from the definition of the derivative, show that the product M (t) := A(t)B(t) is differentiable at t = t 0 and obtain the usual formula for M ′(t 0 ).
Solution [Caution: Usually A(t) and B(t) will not commute.]
M (t 0 + h) − M (t 0 ) h
A(t 0 + h)B(t 0 + h) − A(t 0 )B(t 0 ) h =
[A(t 0 + h) − A(t 0 )] h
B(t 0 + h) + A(t 0 )
[B(t 0 + h) − B(t 0 )] h
Since both A and B are differentiable at t 0 , in the limit as h → 0 we find that M is differentiable at t 0 and M ′(t 0 ) = A′(t 0 )B(t 0 ) + A(t 0 )B′(t 0 ).
B–5. Let ϕn(t) be a sequence of smooth real-valued functions with the properties
(a) ϕn(t) ≥ 0 , (b) ϕn(t) = 0 for |t| ≥ 1 /n, (c)
−∞
ϕn(t) dt = 1.
Note: because of (b), this integral is only over − 1 /n ≤ t ≤ 1 /n. Assume f (x) is uniformly continuous for all x ∈ R and define
fn(x) :=
−∞
f (x − t)ϕn(t) dt.
Show that fn(x) converges uniformly to f (x) for all x ∈ R. [Suggestion: Use f (x) = f (x)
∞ −∞ ϕn(t)^ dt
−∞ f^ (x)ϕn(t)^ dt^.^ Also, note^ explicitly^ where you use the uniform continuity of f ].
Remark: One can show that the approximations fn are also smooth. Thus, this proves that you can approximate a continuous function uniformly on any compact set by a smooth function.
Solution Using the suggestion,
fn(x) − f (x) =
∫ (^1) /n
− 1 /n
[f (x − t) − f (x)] ϕn(t) dt.
Since f is uniformly continuous, given any > 0 there is a δ > 0 so that |f (y) − f (x)| < for any x, y that satisfy |y − x| < δ. Pick some N with 1/N < δ. Then for any n ≥ N , if |t| ≤ 1 /n then |(x − t) − x| ≤ 1 /n < δ so |f (x − t) − f (x)| < for all x ∈ R. Consequently
|fn(x) − f (x)| ≤
|t|≤ 1 /n
ϕn(t) dt = .
Because the right side is independent of x, we have ‖fn − f ‖ ≤ in the uniform norm.
