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24 1 Banach Spaces

1) A Banach space is a complete normed vector space, where completeness is with respect to the metric induced by the norm. 2) Examples of Banach spaces include (Rk, dp) for 1 ≤ p ≤ ∞, where dp is induced by the p-norm, and (C[a,b], d∞) where d∞ is induced by the L∞ norm. 3) The space of continuous functions (C[0,2], k·k1) with the L1 norm is normed but not Banach, as there exists a Cauchy sequence that does not converge.

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536 views4 pages

24 1 Banach Spaces

1) A Banach space is a complete normed vector space, where completeness is with respect to the metric induced by the norm. 2) Examples of Banach spaces include (Rk, dp) for 1 ≤ p ≤ ∞, where dp is induced by the p-norm, and (C[a,b], d∞) where d∞ is induced by the L∞ norm. 3) The space of continuous functions (C[0,2], k·k1) with the L1 norm is normed but not Banach, as there exists a Cauchy sequence that does not converge.

Uploaded by

Dmitri Zaitsev
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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24-1.

Banach spaces
MAU22200 - Advanced Analysis

https://www.maths.tcd.ie/∼zaitsev/Adv-2020
Dmitri Zaitsev zaitsev@maths.tcd.ie

Dmitri Zaitsev (Trinity College Dublin) 24-1. Banach spaces 1/4


Banach space = complete normed space

Definition
A Banach space is a normed vector space (X , k · k) which is complete with
respect to the metric induced by its norm, i.e. d(x, y ) = kx − y k.

Examples
1 We have seen that (Rk , dp ) is complete for every 1 ≤ p ≤ ∞. Since
dp (x, y ) = kx − y kp is induced by the p-norm k · kp , the normed
space (Rk , k · kp ) is Banach.
2 We have also seen that (C [a, b], d∞ ) is complete and again, since
d∞ (f , g ) = supx |f (x) − g (x)| is induced by the L∞ norm
kf k∞ := supx |f (x)|, the normed space (C [a, b], k · k∞ ) is Banach.

Dmitri Zaitsev (Trinity College Dublin) 24-1. Banach spaces 2/4


Space of continuous functions with L1 norm is not Banach
Example
R2
(C [0, 2], k · k1 ) is normed but not Banach,
( where kf k1
( := 0 |f (x)| dx is
xn x < 1 0 x <1
the L1 norm: the sequence fn (x) := → pointwise
1 x ≥1 1 x ≥1
as n → ∞, is Cauchy but divergent in C [0, 2], for otherwise its limit would
be 0 for x < 1 and 1 for x > 1, which cannot be continuous. Indeed,
suppose fn → f in the L1 norm for some f ∈ C [0, 2], i.e.
R2
kfn − f0 k1 = 0 |fn (x) − f (x)|dx → 0, n → ∞.
On each subinterval [0, a] for a < 1,
Ra n
0 |x − f (x)|dx → 0, n → ∞ =⇒ f (x) = 0 for x ∈ [0, a]

Rby2 the uniqueness of limits in metric spaces. Similarly


1 |fn − f (x)|dx → 0, n → ∞ =⇒ f (x) = 1 for x ∈ [1, 2], contradicting
the continuity of f at x = 1.
Dmitri Zaitsev (Trinity College Dublin) 24-1. Banach spaces 3/4
The spaces of sequences `p are Banach
Recall: `p is the normed space of all infinite real sequences x = (xk )k≥1
with kxkp < ∞, together with the `p norm k · kp (as previously defined).
Theorem
For every 1 ≤ p ≤ ∞, the space `p is Banach.
Proof (for p < ∞, the case p = ∞ is analogous). Let (xn ) be a Cauchy
sequence in `p , i.e. xn = (xnk )k≥1 ∈ `p for each n and ∀ε > 0 ∃N
∀m, n ≥ N kxm − xn kp = ( k≥1 |xmk − xnk |p )1/p < ε.
P
(∗)
p 1/p
P
Then, for every fixed j, |xmj − xnj | ≤ ( k≥1 |xmk − xnk | ) < ε proving
that (xnj )n≥1 is Cauchy in R, hence converges to some yj ∈ R, since R is
complete. Collecting all these limits into one sequence y = (yk )k≥1 ,
truncating the sum in (∗) and taking limits as m → ∞, we obtain
( K p 1/p ≤ ε ∀K =⇒ ky − x k ≤ ε
P
k=1 |yk − xnk | ) n p
showing ky kp ≤ ky − xn kp + kxn kp < ∞ =⇒ (1) y ∈ `p and (2) xn → y
in the p-norm as n → ∞, i.e. the Cauchy sequence (xn ) is convergent in `p
as claimed.
Dmitri Zaitsev (Trinity College Dublin) 24-1. Banach spaces 4/4

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