MATH414 FUNCTIONAL ANALYSIS
HOMEWORK 2
Hw 2: 4/70 7/71 10/71 14/71 15/71
Problem (4/70)
(Continuity of vector space operations) Show that in a normed space
X, vector addition and multiplication by scalars are continuous operations with
respect to norm; that is, the mapping dened by (x, y) x+y and (, x) x
are continuous.
Solution
To show continuity of vector addition we have to show that if x
n
x and
y
n
y then x
n
+y
n
x +y. But this follows from
||x
n
+y
n
x y|| ||x
n
x|| +||y
n
y|| 0 + 0 as n .
Similarly to show continuity of scalar multiplication we must show
n
and
x
n
x implies
n
x
n
x. This follows from
||
n
x
n
x|| = ||
n
x
n
n
x +
n
x x||
||
n
x
n
n
x|| +||
n
x x||
= |
n
|
bdd
||x
n
x|| +|
n
| ||x|| 0 + 0.
Remark: If (X, d) and (Y, ) are metric spaces then XY , with product topol-
ogy on it, is also a metric space with metric D given by
D((x, y), (x
, y
)) = d(x, x
) +(y, y
).
So this means that (x
n
, y
n
) (x, y) i x
n
x and y
n
y.
Problem (7/71)
(Absolute convergence) Show that convergence of ||y
1
|| +||y
2
|| +||y
3
|| +
may not imply convergence of y
1
+y
2
+y
3
.
Solution Let F = {(
n
)
n=1
l
| only nitely many terms of (
n
)
n=1
is non-zero}.
Clearly F is a linear subspace of of l
, and so F is a normed space with sup
norm. Consider y
n
F dened by y
n
= (0, 0, ...0,
1
n
2
, 0, ...) where
1
n
2
is n
th
com-
ponent of y
n
. Since ||y
n
|| =
1
n
2
, ||y
1
|| + ||y
2
|| + ||y
3
|| + converges. However
y
1
+y
2
+y
3
converges to (
1
n
2
)
n=1
in l
which is not in F. By uniqueness of
limit we obtain y
1
+y
2
+y
3
doesnt converge in F.
Remark: This is because F is not complete. In fact in a Banach space you
1
cant nd such an example but in an incomplete normed space it is always pos-
sible to nd a sequence as above, so this is a characterization of normed spaces.
Try to prove this fact.
Problem (10/71)
(Schauder basis) Show that if a normed space has a Schauder basis, it is
separable.
Solution Let normed space X has a Schauder basis {e
n
}
n=1
. Lets dene M X
as follows
M = {
N
n=1
n
e
n
: N N and
1
,
2
, ...
N
Q+iQ}.
It is not hard to show M is countable, in fact it is countable union of countable
sets.
M is dense in X. To see this lets x arbitrary x X. By denition ! (
n
)
n=1
in K such that
n=1
n
e
n
= x. Lets x arbitrary > 0. By denition of
convergence N N such that
||
N
n=1
n
e
n
x|| <
2
.
Lets for n = 1, 2, ..., N choose
n
Q+iQ such that
|
n
n
| <
||e
n
||N(N + 1)
.
Clearly
N
n=1
n
e
n
is in M and we have
||
N
n=1
n
e
n
x|| ||
N
n=1
n
e
n
N
n=1
n
e
n
|| +||
N
n=1
n
e
n
x||
<
N
n=1
|
n
n
|||e
n
|| +
2
<
N
n=1
||e
n
||N(N + 1)
||e
n
|| +
2
=
2
+
2
= .
So we can approximate x by the elements of M. Hence M is dense in X.
Remark: The converse of this question is not true, that is, there exists sep-
arable Banach space with no Schauder basis on it.
2
Problem (14/71)
(Quotient space) Let Y be a closed subspace of a normed space (X, ||.||).
Show that ||.||
0
on X/Y is dened by
|| x||
0
= inf
x x
||x||
where x X/Y , that is, x is any coset of Y .
Solution
(N1) Clearly || x||
0
is a non-negative real number for any coset x.
(N2) x =
0 implies 0 x, so || x||
0
= 0. Conversely let || x||
0
= 0. Then
there is a sequence {x
n
} in x such that ||x
n
|| 0. But Y is closed so any coset
of Y must be closed. Hence 0 x. This means that x =
0.
(N3) Notice that ||0 x||
0
= ||
0x||
0
= ||
0||
0
= 0 = 0|| x||
0
. And for any non-
zero in K,
|| x||
0
= || x||
0
= inf
x x
||x|| = inf
x
x
||x|| = || inf
x
x
||
x
|| = |||| x||
0
.
(N4)
|| x + y||
0
= ||
x +y||
0
= inf
z
x+y
||z||
= inf
z x+ y
||z||
= inf
u x, v y
||u +v||
inf
u x, v y
||u|| +||v||
= inf
u x
||u|| + inf
v y
||y|| = || x||
0
+|| y||
0
(Product of normed spaces) If (X
1
, ||.||
1
) and (X
2
, ||.||
2
) are normed spaces,
show that the product vector space X = X
1
X
2
becomes a normed space if
we dene
||x|| = max(||x
1
||
1
, ||x
2
||
2
) [x = (x
1
, x
2
)].
Solution In fact all axioms are easy to verify. Lets, for example, show (N4).
||(x
1
, x
2
) + (y
1
, y
2
)|| = ||(x
1
+y
1
, y
2
+y
2
)||
= max(||x
1
+y
1
||
1
, ||x
2
+y
2
||
2
)
max(||x
1
||
1
+||y
1
||
1
, ||x
2
||
2
+||y
2
||
2
)
max(||x
1
||
1
, ||x
2
||
2
) + max(||y
1
||
1
, ||y
2
||
2
)
= ||(x
1
, x
2
)|| +||(y
1
, y
2
)||.
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