# MA3422 (Functional Analysis 2) Tutorial sheet 8 [March 20, 2015] Name: Solutions ```MA3422 (Functional Analysis 2) Tutorial sheet 8
[March 20, 2015]
Name: Solutions
1. Let E be a normed space and (xn )∞
n=1 a sequence in E, and x ∈ E. Show that the sequence
converges to x in the weak topology if and only if limn→∞ |α(xn −x)| = 0 for each α ∈ E ∗ .
[Hint: an open set for the weak topology that contains x must contain Bq (x, 1) = {y ∈
E : q(x − y) &lt; 1} for some continuous seminorm q on E.]
Solution: Recall first that xn → x as n → ∞ means that for each open U with x ∈ U there
is N such that xn ∈ U for all n ≥ N .
If the sequence converges to x in the weak topology and α ∈ E ∗ then
Bpα (x, ε) = {y ∈ E : pα (x − y) &lt; ε} = {y ∈ E : |α(x − y)| &lt; ε}
is an open set for the weak topology that contains x. So there is N so that
|α(x − xn )| &lt; ε ∀n &gt; N
That means
lim |α(xn − x)| = 0
n→∞
Conversely, assume that this holds for each α ∈ E ∗ . Choose U ⊂ E open in the weak
topology with x ∈ U . Then, as we showed (and according to the hint) there is a continuous
seminorm q (continuous for the weak topology) with Bq (x, 1) ⊆ U . However we also
know that such a seminorm must satisfy
q(x) ≤ C1 pα1 (x) + &middot; &middot; &middot; + Cm pαm (x)
for some m and some α1 , . . . , αm ∈ E ∗ , some constants Cj ≥ 0.
So
q(x − xn ) ≤ C1 pα1 (x − xn ) + &middot; &middot; &middot; + Cm pαm (x − xn ) → 0 as n → ∞
Thus there is N with
q(x − xn ) &lt; 1 ∀n &gt; N
(so xn ∈ Bq (x, 1) ⇒ xn ∈ U ∀n &gt; N ).
Thus xn → x (as n → ∞) in the weak topology.
2. Show that a weakly convergent sequence in a normed space E must be a bounded sequence.
∗
[Hint: If (xn )∞
n=1 is a sequence in E, consider the sequence of operators Tn : E → K
where Tn (α) = α(xn ) and use the uniform boundedness principle. What is the norm of
Tn ∈ E ∗∗ ?]
Solution: Suppose the sequence (xn )∞
n=1 in E, converges to x ∈ E in the weak topology.
By the previous question limn→∞ α(xn −x) = limn→∞ α(xn )−α(x) = 0, or limn→∞ α(xn ) =
∗
α(x). So (α(xn ))∞
n=1 is a convergent sequence and hence bounded (for each α ∈ E ).
Defining (as in the hint) Tn : E ∗ → K by Tn (α) = α(xn ) we actually have Tn = J(xn )
where J : E → E ∗∗ is the canonical embedding. So kTn k = kJ(xn )k = kxn k (and in
particular Tn is bounded for each n). The sequence Tn is pointwise bounded since we have
just observed that Tn (α) = α(xn ) is a bounded sequence for each α ∈ E ∗ .
By the uniform boundedness theorem (which applies since E ∗ is a Banach space), supn kTn k &lt;
∞. So supn kxn k &lt; ∞.
3. Let E = c0 and consider the weak topology on E. Using the fact that E ∗ = `1 , show
that a sequence (xn )∞
n=1 in c0 converges weakly to zero if and only if supn kxn k∞ &lt; ∞
and limn→∞ xn,j = 0 for each j, where xn = (xn,1 , xn,2 , . . .) (and 0 ∈ c0 is the constant
sequence of zeroes).
Solution: By the previous question, if a sequence (xn )∞
n=1 in c0 converges weakly to zero
(that is converges in the weak topology to 0), then supn kxn k∞ &lt; ∞. If we define e∗j ∈ c∗0
by
e∗j (y) = e∗j (y1 , y2 , . . .) = yj
we have limn→∞ e∗j (xn ) = limn→∞ xn,j = e∗j (0) = 0 (using question 1).
For the converse, asssume supn kxn k∞ = M &lt; ∞ and limn→∞ xn,j = 0. Take α ∈ c∗0 . By
the identification c∗0 = `1 there is z = (z1 , z2 , . . .) ∈ `1 with
α(y) = α(y1 , y2 , . . .) =
∞
X
yj zj
j=1
We can estimate
∞
X
|α(xn )| = xn,j zj ≤
j=1
∞
X
|xn,j ||zj |
j=1
=
≤
N
X
|xn,j ||zj | +
j=1
j=N +1
N
X
∞
X
|xn,j ||zj | +
j=1
≤
∞
X
N
X
|xn,j ||zj |
kxn k∞ |zj |
j=N +1
|xn,j ||zj | + M
j=1
∞
X
j=N +1
2
|zj |
P
P∞
Since ∞
|z
|
&lt;
∞,
we
can
choose
N
so
big
that
j
j=1
j=N +1 |zj | is small (say &lt; ε/(2M +
PN
1) for ε &gt; 0 given). Then limn→∞ j=1 |xn,j ||zj | = 0, and so for all n big we will have
PN
j=1 |xn,j ||zj | &lt; ε/2 also. Thus limn→∞ α(xn ) = 0.
As this holds for each α ∈ E ∗ , it shows (by question 1) that the sequence (xn )∞
n=1 converges to 0 in the weak topology.
Richard M. Timoney
3
```