Sequence and Function Spaces

ℓ1,ℓ2,ℓ∞\ell^1, \ell^2, \ell^\infty and C[0,1]C[0,1] — vector spaces under pointwise operations, metric spaces via the norm ∥x−y∥\|x - y\|. A normed space bridges both worlds. Watch the inclusion chain ℓ1⊂ℓ2⊂ℓ∞\ell^1 \subset \ell^2 \subset \ell^\infty through xn=1/nαx_n = 1/n^\alpha. Completeness of ℓp\ell^p and C[0,1]C[0,1] (Banach spaces) is stated as a known theorem only.

ℓ1={x:∑∣xn∣<∞},ℓ2={x:∑∣xn∣2<∞},ℓ∞={x:sup⁡∣xn∣<∞}\ell^1 = \{x : \sum |x_n| < \infty\}, \quad \ell^2 = \{x : \sum |x_n|^2 < \infty\}, \quad \ell^\infty = \{x : \sup |x_n| < \infty\}
∥x∥1=∑∣xn∣,∥x∥2=(∑∣xn∣2)1/2,∥x∥∞=sup⁡n∣xn∣\|x\|_1 = \sum |x_n|, \quad \|x\|_2 = \left(\sum |x_n|^2\right)^{1/2}, \quad \|x\|_\infty = \sup_n |x_n|
ℓ1⊂ℓ2⊂ℓ∞,d(x,y)=∥x−y∥\ell^1 \subset \ell^2 \subset \ell^\infty, \qquad d(x, y) = \|x - y\|

Examples & Exercises

Prove (1,1/2,1/3,… )(1, 1/2, 1/3, \dots) belongs to ℓ2\ell^2 but not to ℓ1\ell^1.

∑n=1∞1n2=π26<∞\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty
∑n=1∞1n=∞\sum_{n=1}^{\infty} \frac{1}{n} = \infty
α=1\alpha = 1 — inside ℓ2\ell^2, outside ℓ1\ell^1.
Show proof
  1. Put xn=1/nx_n = 1/n. Then ∑n=1∞xn2=∑n=1∞1/n2=π2/6<∞\sum_{n=1}^{\infty} x_n^2 = \sum_{n=1}^{\infty} 1/n^2 = \pi^2/6 < \infty: the sum of squares is finite, so x∈ℓ2x \in \ell^2.
  2. But the harmonic series ∑n=1∞∣xn∣=∑n=1∞1/n=∞\sum_{n=1}^{\infty} |x_n| = \sum_{n=1}^{\infty} 1/n = \infty diverges, so x∉ℓ1x \notin \ell^1.
  3. Hence (1,1/2,1/3,… )∈ℓ2∖ℓ1(1, 1/2, 1/3, \dots) \in \ell^2 \setminus \ell^1 — a sequence whose squares are summable but whose absolute values are not. This is the key example for ℓ1⊊ℓ2\ell^1 \subsetneq \ell^2.

Prove ℓ1⊂ℓ2\ell^1 \subset \ell^2.

∑∣xn∣<∞  ⟹  ∑xn2<∞\sum |x_n| < \infty \implies \sum x_n^2 < \infty
α=1.5>1\alpha = 1.5 > 1 — membership in ℓ1\ell^1 forces membership in ℓ2\ell^2.
Show proof
  1. Let x∈ℓ1x \in \ell^1. Since ∑∣xn∣<∞\sum |x_n| < \infty we have ∣xn∣→0|x_n| \to 0, so there is an NN with ∣xn∣<1|x_n| < 1 for all n≥Nn \geq N.
  2. For n≥Nn \geq N, xn2=∣xn∣⋅∣xn∣≤∣xn∣⋅1=∣xn∣x_n^2 = |x_n| \cdot |x_n| \leq |x_n| \cdot 1 = |x_n| — the square of a small number is smaller than the number itself.
  3. Hence ∑n≥Nxn2≤∑n≥N∣xn∣≤∑∣xn∣<∞\sum_{n \geq N} x_n^2 \leq \sum_{n \geq N} |x_n| \leq \sum |x_n| < \infty. The finitely many initial terms do not affect convergence, so ∑xn2<∞\sum x_n^2 < \infty — i.e. x∈ℓ2x \in \ell^2.

Prove d∞(x,y)=sup⁡n∣xn−yn∣d_\infty(x, y) = \sup_n |x_n - y_n| is a metric on ℓ∞\ell^\infty.

d∞(x,y)=sup⁡n∣xn−yn∣d_\infty(x, y) = \sup_n |x_n - y_n|
p=∞p = \infty — the unit ball becomes a square.
Show proof
  1. If x,y∈ℓ∞x, y \in \ell^\infty then (∣xn−yn∣)(|x_n - y_n|) is a bounded sequence, so the supremum exists and is finite — d∞(x,y)d_\infty(x, y) is well defined and d∞(x,y)≥0d_\infty(x, y) \geq 0.
  2. d∞(x,y)=0d_\infty(x, y) = 0 iff ∣xn−yn∣=0|x_n - y_n| = 0 for every nn, iff x=yx = y. Symmetry d∞(x,y)=d∞(y,x)d_\infty(x, y) = d_\infty(y, x) follows from ∣xn−yn∣=∣yn−xn∣|x_n - y_n| = |y_n - x_n|.
  3. Triangle inequality: for each nn, ∣xn−zn∣≤∣xn−yn∣+∣yn−zn∣≤d∞(x,y)+d∞(y,z)|x_n - z_n| \leq |x_n - y_n| + |y_n - z_n| \leq d_\infty(x, y) + d_\infty(y, z). The right side does not depend on nn, so taking suprema gives d∞(x,z)≤d∞(x,y)+d∞(y,z)d_\infty(x, z) \leq d_\infty(x, y) + d_\infty(y, z).
  4. Thus d∞d_\infty satisfies all the metric axioms — ℓ∞\ell^\infty is a metric space, and this is exactly the metric d(x,y)=∥x−y∥d(x, y) = \|x - y\| induced by the norm ∥x∥∞=sup⁡n∣xn∣\|x\|_\infty = \sup_n |x_n|.

2. Sequences