MA3422 (Functional Analysis 2) Tutorial sheet 3
[February 6, 2015]
Name: Solutions
1. For H a Hilbert space and y, z ∈ H, define T : H → H by T (x) = hx, yiz. Verify that
T ∈ B(H) and kT kop = kykkzk.
Solution: To show that T is linear,
T (x1 + λx2 ) = hx1 + λx2 , yiz = (hx1 , yi + λhx2 , yi)z = hx1 , yiz + λhx2 , yiz
= T (x1 ) + λT (x2 )
To show kT kop ≤ kykkzk, we use Cauchy-Schwarz
kT (x)k = |hx, yi|kzk ≤ kxkkykkzk = (kykkzk)kxk
Then if y = 0 we have kT kop = 0. If y 6= 0, take x = y/kyk and we get (since kxk = 1)
|hy, yi|
y
kyk2
=
,
y
z
kzk
==
kzk = kykkz|
kT kop ≤ kT (x)k = kyk
kyk
kyk
Also compute T 2 (x).
Solution:
T 2 (x) = T (T (x)) = hT (x), yiz = hhx, yiz, yiz = hx, yihz, yiz = hz, yiT (x)
2. Let H = Kn be the n-dimensional Hilbert space (with the k·k2 or euclidean norm). For g =
(g1 , g2 , . . . , gn ) ∈ Kn , define Mg ∈ B(Kn ) by Mg (x1 , x2 , . . . , xn ) = (g1 x1 , g2 x2 , . . . , gn xn ).
Show that kMg kop = max1≤i≤n |gi |.
Solution: Recall the notation kgk∞ = max1≤i≤n |gi |. We have
v
uX
u n
kMg (x1 , x2 , . . . , xn )k = t
|gj xj |2
j=1
v
uX
u n
= t
|gj |2 |xj |2
j=1
v
uX
u n
≤ t
kgk2∞ |xj |2
j=1
v
uX
u n
= kgk∞ t
|xj |2
j=1
= kgk∞ k(x1 , x2 , . . . , xn )k2
This shows kMg kop ≤ kgk∞ .
To show the reverse inequality, take x = ej (a standard basis vector for Kn ) and note
that Mg (ej ) = gj ej , kMg (ej )k2 = |gj |kej k2 = |gj | and (since kej k2 = 1) we then have
kMg kop ≥ |gj | for each j (1 ≤ j ≤ n). Thus kMg kop ≥ kgk∞ .
This completes the proof.
3. For H = L2 [0, 1], define Mt : H → H by (Mt f )(t) = tf (t). Show that kMt kop = 1.
[Hint: First check kMt kop ≤ 1 and then choose f supported close to 1.]
Solution: Since |t| ≤ 1 for 0 ≤ t ≤ 1 we have
s
s
Z 1
Z
2
|tf (t)| dt ≤
kMt f k2 =
0
1
|f (t)|2 dt = kf k2
0
and so kMt kop ≤ 1.
If we take f = χ[1−δ,1] (where 0 < δ < 1) then
s
sZ
Z 1
kf k2 =
(χ[1−δ,1] )(t))2 dt =
0
s
Z
√
δ
r
q
1 − (1 − δ)3
t2 dt = [t3 /3]11−δ =
3
1−δ
(tχ[1−δ,1] )(t))2 dt =
kMt f k2 =
1 dt =
1−δ
sZ
1
1
0
1
So
kMt kop
kMt f k2
≥
=
kf k2
r
1 − (1 − δ)3
=
3δ
r
3δ − 3δ 2 + δ 3 p
= 1 − δ + δ 2 /3
3δ
Let δ → 0 to get kMt kop ≥ 1 and conclude the proof.
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Remark: Perhaps it is almost easier to show the more general result that for g ∈ L∞ [0, 1]
the operator Mg : L2 [0, 1] → L2 [0, 1] given by Mg (f ) = gf has kMg kop = kgk∞ . (Q3 was the
special case where g(t) = t.)
For the inequality kMg kop ≤ kgk∞ , use the fact that |g(t)| ≤ kgk∞ for almost all t ∈ [0, 1]
to justify
s
s
s
Z 1
Z 1
Z 1
2
2
2
kMg (f )k2 =
|g(t)f (t)| dt =
|g(t)| |f (t)| dt ≤
kgk2∞ |f (t)|2 dt
0
0
0
s
Z
1
|f (t)|2 dt = kgk∞ kf k2
= kgk∞
0
For the reverse inequality kMg kop ≥ kgk∞ , it is certainly true if kgk∞ = 0. If kgk∞ > 0 then
for 0 < δ < kgk∞ ,
Eδ = {t ∈ [0, 1] : |g(t)| > kgk∞ − δ}
has positive measure (otherwise we would have |g(t)| ≤ kgk∞ − δ almost everywhere, contradicting the definition of the essential supremum kgk∞ ).
We can take f = χEδ and compute
s
sZ
Z 1
p
kf k2 =
χEδ (t)2 dt =
1 dt = µ(Eδ )
0
Eδ
(where µ is Lebesgue measure). Also
s
sZ
Z 1
kMg (f )k2 =
|g(t)|2 χEδ (t)2 dt =
0
sZ
≥
|g(t)|2 1 dt
Eδ
p
(kgk∞ − δ)2 dt = (kgk∞ − δ) µ(Eδ )
Eδ
So
kMg kop ≥
kMg (f )k2
= kgk∞ − δ
kf k2
Let δ → 0 to get kMg kop ≥ kgk∞ .
Combining that with the reverse inequality, we have kMg kop = kgk∞ .
This way we have only actually integrated multiples of characteristic functions. By choosing
f to be supported where g has nearly its maximum modulus, we see that we can’t improve on
the rather easy estimate that gave kMg kop ≤ kgk∞ .
Richard M. Timoney
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