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[MKR&*42]

http://img.buzznet.com/assets/imgx/2...g-21166996.jpg

[Sian]

26308

[Kardan]

Theorem 2.3.6. For any 1 < p < q < 1 we have
l1 ✓ lp ✓ lq ✓ c0 ✓ c ✓ l1,
and
(2.3.4) kxkl1  kxklq  kxklp  kxkl1 ,
for all x 2 l1.
Proof. We begin by proving that l1 ✓ lp. If x 2 l1 then the series
P
1n
=1 xn is absolutely
convergent and there exists n0 such that |xn|  1 for all n + n0. Hence,
Pn0−1
n=1 |xn|p < 1 because it is a finte sum, and
X
n&n0
|xn|p 
X
n&n0
|xn| < 1.
This proves that l1 ✓ lp. Analogously, for 1 < p < q < 1 and x 2 lp we have that
|xn|p  1 (hence |xn|  1) for n + n0 and therefore
X
n&n0
|xn|q 
X
n&n0
|xn|p.
26
It follows that lp ✓ lq. If x 2 lq then the series
P
1n
=1 |xn|q is convergent and therefore
|xn|q ! 0 as n!1. Thismeans that x 2 c0 and proves the inclusion lq ✓ c0. Recall that
every convergent sequence of real numbers is bounded. Hence, the inclusion c0 ✓ c ✓ l1
is trivial.
It remains to prove the chain of inequalities (2.3.4). It is sufficient to prove that
kxklp  kxkl1 for kxkl1 = 1,
kxklq  kxklp for kxklp = 1,
kxkl1  kxklq for kxklq = 1.
(2.3.5)
Indeed, for an arbitrary x 2 l1 we have that x/kxkl1 has l1-norm 1; if we already know
that (2.3.5) holds, then by homogeneity
1
kxkl1 kxklp 
1
kxkl1 kxkl1 .
Analogously the other inequalities can be extended to arbitrary x and (2.3.4) will be valid
for all x 2 l1.
We start by proving the first inequality in (2.3.5). Let x 2 l1 with kxkl1 = 1. Then,
kxklp =
✓ 1X
n=1 |xn|p
◆1
p

✓ 1X
n=1 |xn|
◆1
p
= kxk1/p
l1 = kxkl1 .
The middle inequality holds because if kxkl1 = 1, each entry in x must have absolute
value at most 1.
Next, for x 2 lp with kxklp = 1 we can write
kxkq
lq =
1X
n=1 |xn|q 
1X n=1
|xn|p
=
kxkp
lp .
Thus,
kxklq  kxk
p
q
lp = kxklp
and the second inequality in (2.3.5) holds. Finally, let x 2 lq. Then x is bounded and
converges to zero, so
sup
n |xn| =
✓
sup
n |xn|q
◆1
q

✓ 1X
n=1 |xn|q
◆1
q
.
27

[Bankai90]

May our prays be with MH370.

[MKR&*42]

die sonne scheint noch

oh what

[Tyler]

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[Bankai90]

10 COLOR 63,0
20 CLS
30 C=22
40 X=50
50 Y=50
51 REM ** START OF DRAWING LOOP **
60 IF JOY = 64 THEN X=X+1
70 IF JOY = 256 THEN X=X-1
80 IF JOY = 32 THEN Y=Y+1
90 IF JOY = 128 THEN Y=Y-1
100 IF JOY = 2048 THEN C=C+1
110 PLOT X,Y,C
120 GOTO 51

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