MA244 Analysis III
Solutions. Sheet 4.
NB. THESE ARE JUST SKELETON SOLUTIONS, USE WISELY!
Questions for credit: 8 (5 points), 3 (7 points), 4 (7 points) and 7 (6 points)
0.1
Uniform convergence.
1. (Q.1)
(a) Both series converge uniformly by the Weierstrass M -test: for x ∈ [−M/2, M/2],
∞
X
n=M
∞
X
1
1
≤
< ∞.
2
(n ± x)
(n − M/2)2
n=M
P
1
Therefore, the uniform convergence of the series fM (x) = ∞
n=M (n+x)2 +
P∞
2
1
n=M (n−x)2 on [−M/2, M/2] follows from the M-test with Mn = (n−M/2)2 .
1
1
Each of functions hn (x) := (n−x)
2 + (n+x)2 , n = M, M +1, . . . is continuous
on [-M/2,M/2], hence the limit fM is also continuous as the convergence
0
is uniform.
P∞ h0 n s are continuously differentiable on (−M/2, M/2), the4 seuniformly by the M-test with Mn = (n−M/2)3
ries
n=M hn converges
P∞
0
and the series
n=M hn converges. Therefore, fM is differentiable on
(−M/2, M/2) P
and the derivative can be calculated by term-wise differen0
0
tiation, fM
= ∞
n=M hn .
(b) We need to show that F is defined, continuous and differentiable on (k, k+
1) for every k ∈ Z. Let x ∈ (k, k + 1). Suppose k ≥ 0. Then
2(k+1)
∞
∞
X
X
1
1
1
+
+
.
(x − p)2 p=2k+3 (x − p)2 p=2k+3 (x + p)2
X
F (x) =
p=−2(k+1)
The finite sum in the right hand side of the above is continuous, differentiable for any x ∈ (k, k + 1). The sum of two infinite sums in the right
hand side is continuous differentiable on (−k − 3/2, k + 3/2) by Question
2 with M = 2k + 3. As (k, k + 1) ⊂ (−k − 3/2, k + 3/2), we are done.
The k < 0 case can be treated in exactly the same way. Periodicity: for
any N ∈ Z, x ∈ R \ Z,
F (x + 1) =
=
N
X
lim
N1 →−∞
=
N
2 −1
X
1
1
+ lim
2
N2 →∞
(x − n)
(x − n)2
−1
n=N
N
−1
X
lim
N1 →−∞
n=N1
N2
X
1
1
+
lim
(x + 1 − n)2 N2 →∞ n=N +1 (x + 1 − n)2
n=N1
lim
N1 →−∞
N
−1
X
n=N1
N2
X
1
1
+ lim
2
N2 →∞
(x − n)
(x − n)2
n=N
N
−1
X
∞
X
1
1
=
+
= F (x).
2
2
(x − n)
(x
−
n)
n=−∞
n=N
All of the above operations are justified due to the pointwise convergence
of F on R \ Z.
(c) (i) If g is continuous on R, it is also continuous on [0, 1]. Therefore, it is
bounded on [0, 1] (Analysis II.) Then there exists B ≥ 0: |g(x)| ≤ B for
any x ∈ [0, 1]. Any y ∈ R can be represented in the form y = I + r, where
I is an integer and r ∈ [0, 1). Therefore, |g(y)| = |g(I + r)| = |g(r)| ≤ B,
where the second equality used the periodicity of g. (ii) f is bounded on
R. Therefore, there exists B ≥ 0 such that for any x ∈ R, |f (x)| ≤ B.
Using the functional relation,
|f (x)| ≤
1
1
(|f (x/2)| + |f (x/2 + 1/2)|) ≤ B,
4
2
which is true for any x ∈ R. Repeating the above step n times we find
that for any x ∈ R, |f (x)| ≤ 2−n B → 0 as n → ∞. So f ≡ 0.
(d) (i) Apply L’Hopital’s rule four times:
x2 − sin2 (x)
x→0 x2 sin2 (x)
lim (π 2 cosec2 (πx) − 1/x2 ) = π 2 lim
x→0
2 − 2 cos(2x)
2x − sin(2x)
2
=
π
lim
2
2
x→0 2 sin (x) + 4x sin(2x) + 2x2 cos(2x)
x→0 2x sin (x) + x2 sin(2x)
= π 2 lim
4 sin(2x)
x→0 6 sin(2x) + 12x cos(2x) − 4x2 sin(2x)
= π 2 lim
π2
8 cos(2x)
= .
x→0 12 cos(2x) + 12 cos(2x)
3
= π 2 lim
(ii) Function g is continuous on R \ Z. It is also continuous at x = 0: note
that f1 is continuous on [−1/2, 1/2]. Therefore,
lim g(x) = f1 (0) − lim (π 2 cosec2 (πx) − 1/x2 ) = f1 (0) − π 2 /3 = g(0).
x→0
x→0
As F , cosec2 (π·) are both periodic with period 1 on R \ Z, g is continuous
at every x = n, where n ∈ N:
lim g(x) = lim g(x + n) = lim g(x) = g(0) = g(n).
x→n
x→0
x→0
So g is continuous on R. It is periodic on R \ Z with period 1 and g(n) =
g(0) for any n ∈ Z. Therefore, g is periodic on R with period 1.
(e) (i) As proved in part (d), function g is continuous and periodic. Therefore,
it is bounded by part (c). Consider the restriction of g = F − π 2 cosec2 (π·)
to R \ Z. Then
!
X
1
1
1
1 X
(F (x/2) + F ((x + 1)/2)) =
+
4
4 n∈Z (x/2 − n)2 n∈Z (x/2 + 1/2 − n)2
X
1
1
=
+
2
(x − 2n)
(x − 2n + 1)2
n∈Z
n∈Z
!
X
X
1
=
+
= F (x).
(x − k)2
k∈2Z
k∈2Z+1
!
X
All above steps are justified as the series for F (x) converge absolutely for
x ∈ R \ Z. Also,
1
1
(cosec2 (πx/2) + cosec2 (π(x + 1)/2)) = (cosec2 (πx/2) + sec2 (πx/2))
=
1
= cosec2 (πx/2).
4 sin (πx/2) cos2 (πx/2)
2
Conclusion: both F and cosec2 (π·) satisfy the equation of part (c) on
R \ Z. As the equation is linear, g also satisfies it. As g is continuous at
x = n, n ∈ Z, we can take the limit of x → n, to prove that the equation
of part (c) is satisfied by g for any x ∈ R. As g is bounded, we conclude
that g ≡ 0, which means in particular that for any x ∈ R
X
1
= π 2 cosec2 (πx).
2
(x
−
n)
n∈Z
(ii) For x close to zero, we can rewrite the above as
X
1
1
= π 2 cosec2 (πx) − 2 .
2
2
(x − n)
x
n=1
The left hand side is 2f1 , so it is continuous on [−1/2, 1/2] and the series converges uniformly. Taking the limit x → 0 term-wise is therefore
justified and we get
X 1
= π 2 /6.
2
(n)
n=1
0.2
Norms.
2. (Q.2) (i) || · ||B is not a norm, as it doesn’t separate points: for any non-zero
constant function f ≡ c 6= 0, ||f ||B = ||f 0 ||B = 0, yet f 6= 0. Functions
(1),(3),(4) do define norms on C 1 [0, 1]: (1) Separation of points: ||f ||A = 0 ⇒
||f ||∞ = 0 ⇒ f = 0 as || · ||∞ is a norm; ||f ||C = 0 ⇒ ||f ||∞ = 0 ⇒ f = 0
as || · ||∞ is a norm; ||f ||D = 0 ⇒ f (0) = 0, f 0 ≡ 0. As f 0 ∈ C[0, 1], the
mean value theorem gives that for any x ∈ [0, 1] f (x) = f (0) = 0. (2) The
absolute homogeneity is obvious for all cases. (3) Triangle inequality: all three
candidates have the form ||f || = ||L1 (f )||0 +||L2 (f )||00 , where ||·||0 and ||·||00 are
norms and L1 , L2 are linear operations (Such as differentiation or evaluation
of f at a specific point). Then
||f + g|| = ||L1 (f + g)||0 + ||L2 (f + g)||00 = ||L1 (f ) + L1 (g)||0 + ||L2 (f ) + L2 (g)||00
≤ ||L1 (f )||0 + ||L1 (g)||0 + ||L2 (f )||00 + ||L2 (g)||00 = ||f || + ||g||.
(ii) We claim that there are four equivalence classes: (|| · ||∞ , || · ||A ); (|| · ||1 );
(|| · ||C ), (|| · ||D ). Let’s check it:
• || · ||∞ ∼ || · ||A : for any f ∈ C 1 ([0, 1]),
Z 1
||f ||∞ ≤ ||f ||A = ||f ||∞ +
|f | ≤ ||f ||∞ + ||f ||∞ (1 − 0) = 2||f ||∞ .
0
Therefore, || · ||∞ ∼ || · ||A .
• || · ||C || · ||D : Let fk 7→ xk be a sequence of C 1 ([0, 1])-functions. Then
Z 1
0
||fk ||D = ||fk ||1 = k
xk−1 dx = 1;
0
0
||fk ||C = 1 + ||f ||∞ = 1 + k → ∞ for k → ∞.
Therefore, there is no L > 0 such that for any g ∈ C 1 ([0, 1]), ||g||C ≤
L||g||D and the norms are not equivalent.
• Now, ||·||∞ ||·||1 . Warning: we cannot simply refer to Prop. 38 (Lecture
20) to justify this result, as we are working in a different vector space. So,
1
, n = 1, 2, . . . be a sequence of C 1 -functions. ||fn ||∞ = 1
let fn (x) = 1+nx
R1 1
dx = log(1+n)
→ 0 as n → ∞. Therefore,
for all n’s, but ||fn ||1 = 0 1+nx
n
1
there exists no K > 0 such that for any f ∈ C [0, 1], K||f ||∞ ≤ ||f ||1 . The
non-equivalence is proved. Therefore, || · ||A || · ||1 as A- and ∞-norms
belong to the same equivalence class.
• Similarly, ||·||∞ ||·||C and ||·||1 ||·||C : consider (fn = sin2 (2πn·))n≥1 ⊂
C 1 [0, 1]. Then, ||fn ||∞ = 1, ||fn ||1 = 1/2. But ||fn0 ||∞ = 2πn → ∞ as n →
∞. Therefore, there exist no K1 > 0, K2 > 0 such that ||f ||C ≤ K1 ||f ||∞ ,
||f ||C ≤ K2 ||f ||1 .
• Also, || · ||∞ || · ||D and || · ||1 || · ||D : consider (fn = sin2 (2πn·))n≥1 ⊂
C 1 [0, 1]. Then, ||fn ||∞ = 1, ||fn ||1 = 1/2. But
Z 1
0
||fn ||D = ||fn ||1 = πn
| sin(4πnx)|dx = 2n → ∞
0
as n → ∞. Therefore, there exist no K1 > 0, K2 > 0 such that ||f ||D ≤
K1 ||f ||∞ , ||f ||D ≤ K2 ||f ||1 .
The claim is proved.
3. (Q.3) (a) Let λ = 1/p. Then 1 − λ = 1/q. log(1/x) is convex on (0, ∞). Let
x1 = xp , x2 = y q . The convexity condition reads
1
1
log(1/xp ) + log(1/y q ) ≥ log(1/(xp /p + y q /q)).
p
q
Exponentiating both sides of the above and using that the exponential function
is increasing, we find
1/(xy) ≥ (xp /p + xq /q)−1 ,
which equivalent to (5) as x, y > 0. For x = 0 or y = 0 the inequality (5)
remains valid, as at these special points both sides of (5) turn to zero.
(b) In virtue of (5), at any point x ∈ [a, b], |f (x)||g(x)| ≤ |f (x)|p /p + |g(x)|q /q.
In both sides of the inequality we have continuous functions. Using the theorem
on integral bounds, we get:
Z b
Z b
Z b
p
|f g| ≤
|f | /p +
|g|q /q,
a
a
a
Rb
which holds for any f, g ∈ C[a, b]. If F, G ∈ C[a, b]:
Z
a
|F |p =
Rb
a
|G|q = 1,
b
|F G| ≤ 1/p + 1/q = 1.
a
Rb
Rb
(c) Assume that f, g are not identically zero on [a, b]. Then a |f |p > 0, a |g|q >
Rb
Rb
Rb
Rb
0. Let F = f /( a |f |p )1/p , G = g/( a |g|q )1/q . Clearly, a |F | = 1, a |G| = 1.
Rb
As we proved in part (b), a |F G| ≤ 1. Re-writing this in terms of f, g again
we get the desired inequality:
Z
b
Z
|f g| ≤
a
b
p
1/p Z
|f |
a
b
q
|g|
a
1/q
.
We proved the inequality for any f, g which are not identically zero on [a, b].
It is easy to check that it is also valid for f ≡ 0 or g ≡ 0. For p = q = 2,
Z
b
Z
b
b
Z
|f g| ≤
fg ≤
f
a
a
2
1/2 Z
b
g
2
1/2
.
a
a
4. (Q.4) (a) Assume that f is not identically zero. Choose α = p − 1. Then
q −1 = 1 − p−1 and
b
Z
b
Z
p
q(p−1)
|f | ≤ A
v
1−p−1
|f |
a
.
a
R
But q(p − 1) = p. Dividing both sides with
Z
b
p
b
a
q(p−1)
|f |
1−p−1
> 0 we find
1/p
≤ A,
|f |
a
as required. A quick inspection shows that the above inequality is valid for
f ≡ 0, thus completing the proof.
(b) Applying Holder’s inequality we find that for any g ∈ C[a, b],
Z
b
b
Z
q
|g|
|(f + h)g| ≤ A
R
b
a
Z
|f |p
1/p
b
p
|f + h|
+
,
a
a
where A =
1/q
R
b
a
|h|1/p
1/p
1/p
. By the inequality proved in part (a),
Z
p
|f |
≤A=
a
b
a
1/p
Z
b
1/p
|h|
+
1/p
.
a
Minkowski’s inequality is proved.
R
1/p
b
(c) Let ||f ||p = a |f |p
. Minkowski’s inequality means that the function
|| · ||p : C[a, b] → R satisfies the triangle inequality. The absolute homogeneity
is a doddle to prove. The separation of points: if ||f ||p = 0, then |f |p ≡ 0, as
|f |p is continuous. (See the notes for the proof of p = 1 case.) Therefore f ≡ 0
and we conclude that || · ||p is indeed a norm for any p ≥ 1.
0.3
Completeness.
5. (Q.5) In Question 2 we identified four equivalence classes of norms. Using the
statement quoted in the hint, it is enough to check completeness for just one
representative of each class:
(1) The space (C 1 [0, 1], || · ||∞ ) is not complete: the limit of a uniformly Cauchy
sequence of C 1 -functions has to be continuous, but it doesn’t
have to be differp
entiable. An example was given in the lectures: fn (x) = (x − 1/2)2 + 1/n, n =
1, 2, . . .. (Full details omitted.) By equivalence, (C 1 [0, 1], ||·||A ) is not complete
either.
(2) The space (C 1 [0, 1], ||·||1 ) is not complete. The sequence (fn (x) = tanh(n(x−
1/2))n≥1 ⊂ C 1 [0, 1] is Cauchy w. r. t. || · ||1 (sketch a picture):
Z
1
||fn+k − fn ||1 =
| tanh((n + k)(x − 1/2)) − tanh(n(x − 1/2))|
0
Z
√
1/2−1/ n
Z
+
=
0
√
1/2+1/ n
√
1/2−1/ n
+
!
1
Z
| tanh((n+k)(x−1/2))−tanh(n(x−1/2))|
√
1/2+1/ n
:= I1 + I2 + I3 .
√
But I2 ≤ 4/ n → 0 as n → ∞;
Z 1
I3 =
| tanh((n + k)(x − 1/2)) − tanh(n(x − 1/2))|
√
1/2+1/ n
MV T
=
Z
1
√
1/2+1/ n
1
|k(x − 1/2)|,
cosh ξ(n, k, x)
2
√
where ξ(n, k, x) ∈ (n(x − 1/2), (n + k)(x − 1/2)). Therefore, ξ(n, k, t) ≥ n →
∞ as n → ∞. So, I3 → 0 as n → ∞. Similarly, I1 → 0 as n → ∞. Therefore,
(fn ) is || · ||1 Cauchy. However, the sequence (fn ) does not have a limit in
C 1 [0, 1]. In fact, an argument similar to the one used in the lectures, shows
that (fn ) cannot even converge to a continuous function on [0, 1].
(3) The space (C 1 [0, 1], || · ||C ) is Banach. Really, any sequence (fn )n≥1 ⊂
C 1 [0, 1] which is Cauchy with respect to || · ||D must have the following properties: (i) (fn ) is uniformly Cauchy; (ii) (fn0 ) is uniformly Cauchy. Therefore,
we are dealing with a pointwise convergent series of C 1 -functions such that the
series of derivatives converges uniformly. The limit must be C 1 by Theorem
25. This shows that any Cauchy sequence in (C 1 [0, 1], || · ||C ) converges, which
proves completeness.
(4) The space (C 1 [0, 1], || · ||D ) is not Banach. Let gn ∈ C 1 [0, 1] be the following
sequence: gn (0) = 0 and gn0 (x) = tanh(n(x − 1/2)). Notice that ||gn ||D = ||gn0 ||1
and we know from part (2) that the sequence of gn is Cauchy with respect to
|| · ||D . We also know from part (2) that the sequence (gn0 ) does not converge to any continuous function on [0, 1]. So, the sequence of (gn ) does
not converge to any differentiable
function: let L ∈ C 1 [0, 1]. Then there is
Rx
l ∈ C[0, 1] : L(x) = L(0) + 0 l(t)dt. Therefore, ||gn − L||D = |L(0)| + ||gn0 − l||1
- doesn’t converge to zero as n → ∞.
6. (Q.6) If (an )n≥1 , (bn )≥1 ∈ s00 , there exists N ∈ N: for any n > N , an = bn = 0.
Therefore cn := an +bn = 0, dn := λan = 0 for n > N and any λ ∈ R. Therefore
(cn )n≥1 , (dn )n≥1 ∈ s00 . Therefore, s00 is closed under component-wise addition
and multiplication by scalars. One can also check that these operations satisfy
all linear space axioms, implying that s00 is a linear space over R.
Consider the following sequence of sequences: ( 21n I{n ≤ N })n≥1 ⊂ s00 , N =
1, 2, . . .. Here I{n ≤ N } = 1 if n ≤ N and is zero otherwise. This sequence is
Cauchy:
N +M
||(a)n≥1
−
(a)N
n≥1 ||1
=
NX
+M
k=N +1
−k
2
≤
∞
X
k=N +1
2−k → 0, N → ∞.
However, there is no sequence in s00 to which (an )N
n≥1 converges: consider an
arbitrary (ln )n≥1 ∈ s00 . Then there exists C ∈ N such that ln = 0 for any
n ≥ C. Therefore for any N > C,
−C
||(an )N
9 0 as N → ∞.
n≥1 − (ln )≥1 ||1 ≥ 2
7. (Q.7) Consider the following sequence of piece-wise linear functions (already
used in the lectures to show that (C[a, b]), || · ||1 is not complete): (fn )n≥1 ⊂
C[a, b], fn (x) = −1 for x ≤ (a + b)/2 − (b − a)/(2n), fn (x) = 1 for x ≥
(a + b)/2 + (b − a)/(2n),
fn (x) = 2n
(x − (a + b)/2)
(b − a)
for any other x ∈ [a, b]. For any m, n ∈ N,
Z
(a+b)/2+(a−b)/(2n)
||fn+m −fn ||p ≤
!1/p
2p dx
≤ 2((a−b)/n)1/p → 0 as n → ∞.
(a+b)/2−(a−b)/(2n)
Therefore, the sequence of functions (fn )n≥1 is Cauchy with respect to || · ||p .
The fact that the sequence cannot converge to a continuous function can be
proved either by repeating the argument given in the lectures or as follows: let
l ∈ C[a, b]. It is continuous at the midpoint m = (a + b)/2. Then for any > 0,
there is δ > 0: for any x ∈ (m − δ, m + δ), |f (x) − f (m)| < . Then for any n:
(b − 2)/(2n) < δ/2,
||fn − l||p
1/p
δ
p
p
min(
inf
|1 + x| ,
inf
|1 − x| )
>0
≥
x∈(f (m)−,f (m)+)
x∈(f (m)−,f (m)+)
2
for < 1/2, as no interval of length 2 < 1 can contain both +1 and −1.
Notice that the lower bound we derived does not depend on n. Therefore,
||fn − l|| 9 0 as n → ∞. We found a Cauchy sequence, which does not
converge in (C[a, b], || · ||p ). Therefore, Lp is not complete for any p ≥ 1.
8. (Q.8) (i) Let x ∈ X : f (x) = x be the unique fixed point of f . Applying
f ◦ g = g ◦ f to x we get f (g(x)) = g(f (x)) = g(x). Therefore, g(x) is a fixed
point of f . As the fixed point is unique, we must have g(x) = x, meaning that
x is also a fixed point of g. Yes, g can have more than one fixed point: let
g = Id, the identity map. Then every point of x ∈ X is a fixed point of X
(g(x) = Id(x) = x) and f ◦ g = f ◦ Id = f = Id ◦ f = g ◦ f . (ii) Suppose
f : R → R is the identity map and g : x ∈ R 7→ 1 + |x|. Then f leaves every
point of R fixed, f ◦ g = g ◦ f , but g has no fixed points.
9. (Q.9) x + e−x ≥ 0 ∀x ≥ 1 so f maps U = [1, ∞) into itself. If x 6= y, by the
Mean Value Theorem |f (x) − f (y)| = |f 0 (ξ)| |x − y| for some ξ between x and
y; now |f 0 (ξ)| = |1 − e−ξ | < 1 for any ξ ∈ [1, ∞) so |f (x) − f (y)| < |x − y|.
f has no fixed point since f (x) = x ⇐⇒ 1 + e−x = 0 which is impossible
for x ∈ R and hence in U . The Contraction Mapping Theorem does not apply
because there exists no contraction factor K < 1 such that |f (x) − f (y)| <
1
K|x − y| ∀x 6= y ∈ U . Indeed given K < 1 take x, y and hence ξ > log 1−K
then 1 − e−ξ > K so |f (x) − f (y)| = |f 0 (ξ)| |x − y| > K|x − y|.
0.4
Closed and open sets, continuity.
10. (Q.10) U ∈ U if, ∀x ∈ U, ∃ε > 0 s.t. B(x, ε) ⊂ U . (i) is easy, because 6 ∃x ∈ ∅
and any B(x, ε)
T⊂V.
(ii) For x ∈ nj=1 Uj , ∀j ∈ {1, . . . , n}∃ εj T> 0 with B(x, εj ) ⊂ Uj so, for
ε := min{εj :S
1 ≤ j ≤ n} we have B(x, ε) ⊂ nj=1 Uj .
S
(iii) For x ∈ j∈J Uj , ∃ j ∈ J s.t. x ∈ Uj and ∃ ε s.t. B(x, ε) ⊂ Uj ⊂ j∈J Uj
as required.
11. (Q.11) Follows from f −1 (W \A) = V \f −1 (A) for any A ⊂ W and W \A is open
iff A is closed.
12. (Q.12) (i) For any x, z ∈ V ,
d(x + z, E) = inf ||x + z − y|| ≤ inf (||x − y|| + ||z||) = d(x, E) + ||z||,
y∈E
y∈E
d(x, E) = inf ||x + z − y − z|| ≤ inf (||x + z − y|| + ||z||) = d(x + z, E) + ||z||,
y∈E
y∈E
So,
|d(x + z, E) − d(x, E)| ≤ ||z||.
Therefore, the map d(·, E) is continuous at every x ∈ E by the second definition of continuity (Definition 140 of the notes) with δ = .
(ii) (a) ⇐ (easy direction): Suppose K = L. For any k ∈ L, d(k, L) =
inf y∈L d(k, L) = 0, (the infimum is reached at y = k). Therefore,
ρ(L, L) = max(sup d(k, L), sup d(l, L)) = sup d(k, L) = sup(0) = 0.
k∈L
l∈L
k∈L
k∈L
⇒ (a harder direction): Suppose ρ(K, L) = 0. Then supk∈K d(k, L) = 0 and
supl∈L d(l, K) = 0. Consider the first condition. It implies that for any k ∈ K,
inf l∈L ||k − l || = 0. Consider any null sequence of positive numbers (δn )n≥1 . As
inf l∈L ||k − l || = 0, for any δn > 0 there is ln ∈ L : ||k − ln || ≤ δk . The sequence
(δk )k≥1 converges to k. As L is closed, this implies that k ∈ L. (Proposition
45 of the notes). We conclude that K ⊂ L. Analysing the second condition
supl∈L d(l, K) = 0 in exactly the same way we arrive at L ⊂ K. Therefore
K = L.
(b) This is obvious as max(x, y) = max(y, x).
(c) For any m ∈ M , k ∈ K
d(k, L) := inf ||k − l || ≤ inf (||k − m)|| + ||m − l ||).
l∈L
l∈L
Taking inf m∈M of both sides of the above, we get
d(k, L) ≤ d(k, M ) + inf d(m, L) ≤ d(k, M ) + sup d(m, L).
m∈M
m∈M
Similarly, for any l ∈ L,
d(l, K) ≤ d(l, M ) + sup d(m, K).
m∈M
Substituting the last two inequalities in the expression for ρ we get
ρ(K, L) ≤ max(sup d(k, M ) + sup d(m, L), sup d(l, M ) + sup d(m, K))
k∈K
m∈M
l∈L
m∈M
≤ max(sup d(k, M ), sup d(m, K)) + max( sup d(m, L), sup d(l, M ))
k∈K
m∈M
m∈M
l∈L
= ρ(K, M ) + ρ(M, L).
At the last step we used the magic formula from the hint.
November the 23rd, 2015
Sergey Nazarenko and Oleg Zaboronski.