Functional Analysis Exercise sheet 1 — Solutions
1. We use that
N
X
|xn + yn |p ≤
n=1
N
X
(2 max(|xn |, |yn |))p ≤ 2p
n=1
N
X
(|xn |p + |yn |p ).
n=1
Hence, if x = (xn )n∈N and y = (yn )n∈N are in `p , then also
∞
X
|xn + yn |p < ∞.
n=1
This proves that `p is closed under addition.
Also for α ∈ F,
∞
X
|αxn |p = |α|p
n=1
∞
X
|xn |p .
n=1
Hence, if x ∈ `p , then αx ∈ `p too.
2. The property x ∈ `p is equivalent to
converges if and only if p > 2.
P∞
1
n=1 np/2
< ∞. This series
3. We have
kx + yk2 = hx, xi + 2Re hx, yi + hy, yi ,
kx − yk2 = hx, xi − 2Re hx, yi + hy, yi ,
kx + iyk2 = hx, xi + 2Re hx, iyi + hiy, iyi = hx, xi + 2Im hx, yi − hy, yi ,
kx − iyk2 = hx, xi − 2Re hx, iyi + hiy, iyi = hx, xi − 2Im hx, yi − hy, yi .
Adding these equalities, we deduce the formula.
4. We have
kx + yk2 = hx, xi + 2Re hx, yi + hy, yi ,
kx − yk2 = hx, xi − 2Re hx, yi + hy, yi .
Adding these equalities, we deduce the formula. The equality says
that the sum of the squares of sides of a parallelogram is equal to the
sum of the squares of diagonals.
5. kx + yk2 = hx, xi + 2Re hx, yi + hy, yi = kxk2 + kyk2 .
6. We have
| hxn , yn i − hx, yi | ≤ | hxn , yn i − hx, yn i | + | hx, yn i − hx, yi |
= | hxn − x, yn i | + | hx, yn − yi |
≤ kxn − xkkyn k + kxkkyn − yk
1
Since yn → y, kyn k ≤ kyn − yk + kyk ≤ 1 + kyk for all sufficiently large
n. Hence, for all sufficiently large n,
| hxn , yn i − hx, yi | ≤ kxn − xk(1 + kyk) + kxkkyn − yk → 0.
This proves the claim.
p
7. Consider the function φ(t) = tb − tp on [0, ∞]. Since φ0 (t) = b − tp−1 ,
we have φ0 > 0 for t < b1/(p−1) , and φ0 < 0 for t > b1/(p−1) . Hence, φ
achieves its maximal value at t = b1/(p−1) . This maximal value is
φ(b1/(p−1) ) = b1+1/(p−1) −
bp/(p−1)
1
bq
= (1 − )bp/(p−1) = .
p
p
q
This implies the Young inequality.
2