Sequence and Function Spaces

Whole sequences get collected into a single space here. Each sequence becomes one point of the space.

Sequences can be added and scaled, so the space is a vector space. The distance between two sequences can also be measured. The measuring rule decides which space a sequence fits into. A finite sum of absolute values fits the smallest space. A finite sum of squares fits the middle one. Merely being bounded fits the largest one. So each smaller space sits inside the bigger ones. The norm behind the distance bridges the two worlds. That these spaces are complete is stated as a known theorem only.

For example, picture three nested πŸ“¦ boxes. Anything inside the small box also lies in the middle and the large one. Sequence spaces nest in exactly the same way.

β„“1={x:βˆ‘βˆ£xn∣<∞},β„“2={x:βˆ‘βˆ£xn∣2<∞},β„“βˆž={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\}
βˆ₯xβˆ₯1=βˆ‘βˆ£xn∣,βˆ₯xβˆ₯2=(βˆ‘βˆ£xn∣2)1/2,βˆ₯xβˆ₯∞=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|
β„“1βŠ‚β„“2βŠ‚β„“βˆž,d(x,y)=βˆ₯xβˆ’yβˆ₯\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 β„“2\ell^2 but not to β„“1\ell^1.

βˆ‘n=1∞1n2=Ο€26<∞\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty
βˆ‘n=1∞1n=∞\sum_{n=1}^{\infty} \frac{1}{n} = \infty
Ξ±=1\alpha = 1 β€” inside β„“2\ell^2, outside β„“1\ell^1.
Show proof

  1. Put xn=1/nx_n = 1/n. Then βˆ‘n=1∞xn2=βˆ‘n=1∞1/n2=Ο€2/6<∞\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 xβˆˆβ„“2x \in \ell^2.

    (The value βˆ‘1/n2=Ο€2/6\sum 1/n^2 = \pi^2/6 is proved by Parseval in the muntz-szasz scene; convergence alone follows from the integral test in the series chapter.)

  2. But the harmonic series βˆ‘n=1∞∣xn∣=βˆ‘n=1∞1/n=∞\sum_{n=1}^{\infty} |x_n| = \sum_{n=1}^{\infty} 1/n = \infty diverges, so xβˆ‰β„“1x \notin \ell^1. (Divergence is Oresme's grouping argument: the block 1/(2k+1)+β‹―+1/2k+11/(2^k+1) + \cdots + 1/2^{k+1} has 2k2^k terms each β‰₯1/2k+1\ge 1/2^{k+1}, hence sums to β‰₯1/2\ge 1/2; proved in the series chapter.)

  3. Hence (1,1/2,1/3,… )βˆˆβ„“2βˆ–β„“1(1, 1/2, 1/3, \dots) \in \ell^2 \setminus \ell^1 β€” a sequence whose squares are summable but whose absolute values are not. This is the key example for β„“1βŠŠβ„“2\ell^1 \subsetneq \ell^2.

Prove β„“1βŠ‚β„“2\ell^1 \subset \ell^2.

βˆ‘βˆ£xn∣<βˆžβ€…β€ŠβŸΉβ€…β€Šβˆ‘xn2<∞\sum |x_n| < \infty \implies \sum x_n^2 < \infty
Ξ±=1.5>1\alpha = 1.5 > 1 β€” membership in β„“1\ell^1 forces membership in β„“2\ell^2.
Show proof

  1. Let xβˆˆβ„“1x \in \ell^1. Since βˆ‘βˆ£xn∣<∞\sum |x_n| < \infty we have ∣xnβˆ£β†’0|x_n| \to 0, so there is an NN with ∣xn∣<1|x_n| < 1 for all nβ‰₯Nn \geq N.

  2. For nβ‰₯Nn \geq N, xn2=∣xnβˆ£β‹…βˆ£xnβˆ£β‰€βˆ£xnβˆ£β‹…1=∣xn∣x_n^2 = |x_n| \cdot |x_n| \leq |x_n| \cdot 1 = |x_n| β€” the square of a small number is smaller than the number itself.

  3. Hence βˆ‘nβ‰₯Nxn2β‰€βˆ‘nβ‰₯N∣xnβˆ£β‰€βˆ‘βˆ£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 βˆ‘xn2<∞\sum x_n^2 < \infty β€” i.e. xβˆˆβ„“2x \in \ell^2.

Prove d∞(x,y)=sup⁑n∣xnβˆ’yn∣d_\infty(x, y) = \sup_n |x_n - y_n| is a metric on β„“βˆž\ell^\infty.

d∞(x,y)=sup⁑n∣xnβˆ’yn∣d_\infty(x, y) = \sup_n |x_n - y_n|
p=∞p = \infty β€” the unit ball becomes a square.
Show proof

  1. If x,yβˆˆβ„“βˆžx, y \in \ell^\infty then for every nn, ∣xnβˆ’ynβˆ£β‰€βˆ£xn∣+∣ynβˆ£β‰€βˆ₯xβˆ₯∞+βˆ₯yβˆ₯∞<∞|x_n - y_n| \le |x_n| + |y_n| \le \|x\|_\infty + \|y\|_\infty < \infty, so (∣xnβˆ’yn∣)(|x_n - y_n|) is bounded by βˆ₯xβˆ₯∞+βˆ₯yβˆ₯∞\|x\|_\infty + \|y\|_\infty and the supremum exists and is finite β€” d∞(x,y)d_\infty(x, y) is well defined and d∞(x,y)β‰₯0d_\infty(x, y) \geq 0.

  2. d∞(x,y)=0d_\infty(x, y) = 0 iff ∣xnβˆ’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 ∣xnβˆ’yn∣=∣ynβˆ’xn∣|x_n - y_n| = |y_n - x_n|.

  3. Triangle inequality: for each nn, ∣xnβˆ’znβˆ£β‰€βˆ£xnβˆ’yn∣+∣ynβˆ’znβˆ£β‰€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 β€” β„“βˆž\ell^\infty is a metric space, and this is exactly the metric d(x,y)=βˆ₯xβˆ’yβˆ₯d(x, y) = \|x - y\| induced by the norm βˆ₯xβˆ₯∞=sup⁑n∣xn∣\|x\|_\infty = \sup_n |x_n|.

3. Sequences