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WA 8: Solutions
- Problem 6.9 (see also Problem 2.7 for the definition of the space c 0 ).
Solution. Given x ∈ `
1 , define the linear functional Fx on c 0 by
Fx(y) =
∞ ∑
k=
xkyk.
This functional is well defined and bounded:
|Fx(y)| =
∞ ∑
k=
xkyk
∞ ∑
k=
|xkyk| ≤ sup k∈N
|yk|
∞ ∑
k=
|xk| = ‖x‖ 1 ‖y‖∞,
and ‖Fx‖ ≤ ‖x‖ 1. On the other hand, define the sequence {y
(n) } in c 0 as follows
y
(n) k
xk |xk | , xk 6 = 0, 1 ≤ k ≤ n,
0 , xk = 0, 1 ≤ k ≤ n,
0 , k > n.
Clearly, ‖y
(n) ‖∞ = 1, and
|Fx(y
(n) )| =
n ∑
k=
|xk| →
∞ ∑
k=
|xk| = ‖x‖ 1
as n → ∞. Therefore, ‖Fx‖ = ‖x‖ 1. The mapping T : ` 1 → (c 0 )
∗ defined by
T x = Fx, which is clearly linear, is therefore norm preserving.
Now we show that T is surjective. Let g ∈ (c 0 )
∗ be arbitrary. Set xk = g(ek),
k ∈ N, where ek is a sequence in c 0 with 1 on the k-th position and 0 elsewhere.
Define y
(n) as above. Then
g(y
(n) ) =
n ∑
k=
|xk| ≤ ‖g‖.
Therefore, the series
k= |xk| converges, the sequence x = {xk} belongs to `
1 ,
and ‖x‖ 1 ≤ ‖g‖. (In fact, ‖x‖ 1 = ‖g‖ by the same argument as in the preceding
paragraph.) If y ∈ c 0 then
∥ ∥ ∥ ∥ ∥
y −
n ∑
k=
ykek
= sup i>n
|yi| → 0
as n → ∞. Therefore,
g(y) = lim n→∞
n ∑
k=
ykg(ek) = lim n→∞
n ∑
k=
xkyk =
∞ ∑
k=
xkyk = Fx(y),
i.e., g = Fx.
Finally, we obtain that T is a norm preserving isomorphism of Banach spaces `
1
and (c 0 )
∗ .
The argument above also shows that every vector x ∈ `
1 determines a bounded
linear functional on `
∞
. However, the analogue of the mapping T above for (c 0 )
∗
1
replaced with (`
∞ )
∗ is, in fact, not surjective as a mapping from `
1 to (`
∞ )
∗
. At
least, the argument above fails because
∥ ∥ ∥ ∥ ∥
y −
n ∑
k=
ykek
= sup i>n
|yi| → 0
as n → ∞ is not longer true for y ∈ `
∞ unless y ∈ c 0.
- Exercise 7.3, page 71
Solution. For an arbitrary x ∈ E one obtains by the triangle inequality and
the Cauchy-Schwarz inequality that
‖Ax‖ =
A
n ∑
k=
〈x, ek〉ek
n ∑
k=
|〈x, ek〉| · ‖Aek‖
n ∑
k=
|〈x, ek〉|
2
n ∑
k=
‖Aek‖
2
n ∑
k=
‖Aek‖
2
‖x‖.
Therefore, A is bounded and ‖A‖ ≤
n k= ‖Aek‖ 2
- Exercise 7.12, page 75
Solution. By the Cauchy–Schwarz inequality,
‖Kf ‖
2
1
0
t
0
(t − s)f (s) ds
2
dt
1
0
t
0
(t − s)
2 ds
t
0
|f (s)|
2 ds
dt
1
0
t 3
dt
1
0
|f (s)|
2 ds =
‖f ‖
2 .
Therefore, ‖K‖ ≤ 1 /
We will now show by induction on n that
(K
n f )(t) =
∫ (^) t
0
(t − s) 2 n− 1
(2n − 1)!
f (s) ds.
Of course, the definition of K serves as an induction base (for n = 1). Suppose now
that the formula is true for some n. We will show that the formula is true then for
n + 1. We have
(K
n+ f )(t) =
t
0
(t − s)(K
n f )(s) ds =
t
0
(t − s)
s
0
(s − τ )
2 n− 1
(2n − 1)!
f (τ ) dτ
ds
t
0
f (τ )
t
τ
(t − s)
(s − τ )
2 n− 1
(2n − 1)!
ds
dτ.
The change of the integration order in the repeated integral above is legitimate due
to Fubini’s theorem. There is also a longer way to justify this, bypassing Fubini’s
theorem. In the assumption that f is continuous it is legitimate as the integrand
This definition is correct because
∥ ∥ ∥ ∥ ∥
∞ ∑
k=
λkxkek
2
∞ ∑
k=
|λk|
2 |xk|
2 ≤ sup n∈N
|λn|
2
∞ ∑
k=
|xk|
2 = sup n∈N
|λn|
2 ‖x‖
2 ,
and by the completeness of the orthonormal sequence {en} the series
k= λkxkek
converges. Clearly D is a linear operator. Moreover, in this case D is bounded and
‖D‖ ≤ sup n∈N |λn|.
Finally, we obtain that there exists a bounded linear operator D on a Hilbert
space H such that Den = λnen for all n ∈ N if and only if the sequence {λn} is
bounded, and in this case ‖D‖ = supn∈N |λn|.
- Problem 7.
Solution. First observe that y(t) ≡ 1 does not belong to the range of M.
Indeed, otherwise x(t) = y(t)/t = 1/t would belong to L 2 [0, 1], which is not the
case. Thus, the range of M is not the whole space L
2 [0, 1]. Let us show that this
range is dense in L
2 [0, 1]. Given f ∈ L
2 [0, 1], define
fn(t) =
f (t)/t, t ∈ [1/n, 1]
0 , t ∈ [0, 1 /n).
Since
∫ (^1)
0
|fn(t)|
2 dt =
1 /n
|f (t)|
2
t 2
dt ≤ n
2
1 /n
|f (t)|
2 dt ≤ n
2
0
|f (t)|
2 dt = n
2 ‖f ‖
2 < ∞,
we have fn ∈ L 2 [0, 1]. Since
‖f − Dfn‖
2
1
0
|f (t) − tfn(t)|
2 dt =
1 /n
0
|f (t)|
2 dt → 0
as n → ∞ by the absolute continuity of Lebesgue integral, and since f is arbitrary,
the range of D is dense in L
2 [0, 1], i.e., its closure is the whole space L
2 [0, 1].
- Problem 7.
Solution. The linearity of V is obvious. We have, for an arbitrary x ∈ L
2 [0, 1],
by the Cauchy–Schwarz inequality
‖V x‖
2
0
∫ (^) t
0
x(s) ds
2
dt ≤
0
t
0
ds
t
0
|x(s)|
2 ds
dt
0
t
1
0
|x(s)|
2 ds
dt = ‖x‖
2
0
t dt =
‖x‖
2 .
Therefore, V is bounded and ‖V ‖ ≤
1 √ 2
- Problem 7.
Solution. Let ψ be a nonzero vector in the range of T. Then for every x ∈ H one
has T x = α(x)ψ with some α(x) ∈ C. Since for any x, y ∈ H and μ, ν ∈ C one has
T (μx+νy) = α(μx+νy)ψ and μT x+νT y = μα(x)ψ +να(y)ψ = (μα(x)+να(y))ψ,
and T is linear, we obtain that α(μx + νy) = μα(x) + να(y), i.e., α is a linear
functional on H. Since T is a bounded operator,
‖α‖ = sup ‖x‖=
|α(x)| =
‖ψ‖
sup ‖x‖=
‖T x‖ =
‖T ‖
‖ψ‖
the functional α is bounded. By the Riesz–Frech´et theorem, there exists a vector
φ in H such that α(x) = 〈x, φ〉 for every x ∈ H, and ‖α‖ = ‖φ‖. Therefore,
T x = 〈x, φ〉ψ and ‖φ‖ = ‖T ‖/‖ψ‖, i.e., ‖T ‖ = ‖φ‖ ‖ψ‖.
