A two sided seriesP∞ n=−∞vnin (V, k·k) is said to be absolutely convergent ifP∞ n=−∞kvnk is convergent in R

1. Two sided Series

Let F be the field of real numbers or the field of complex numbers. All the normed spaces are assumed to be over F.

Let (V, k · k) be a normed space. A two sided sequence on V is a function v : Z → V. A two sided sequence v : Z → V is also denoted by (vn)^∞_n=−∞. For each n ≥ 0, we define a vector

sn= X

|k|≤n

v_k in V . The Eisenstein sumP∞

n=−∞vn of v is defined to be the limit of the sequence (sn) in V i.e.

∞

X

n=−∞

vn= lim

n→∞sn

when (s_n) is convergent in (V, k · k).

Definition 1.1. A two sided seriesP∞

n=−∞vnin (V, k·k) is said to be absolutely convergent ifP∞

n=−∞kv_nk is convergent in R.

Proposition 1.1. Let (V, k · k) be a Banach space. If a seriesP∞

n=−∞v_nin V is absolutely convergent, then it is convergent in (V, k · k).

Proof. Let t_n=P

|k|≤nkv_kk for n ≥ 0. Then (t_n) is convergent in R by assumption. Hence (tn) is a Cauchy sequence in R. For any  > 0, there exists N ∈ N so that

|t_n− t_m| <  whenever n, m ≥ N_. By triangle inequality, for any n > m ≥ N,

ks_n− s_mk =

X

m+1≤|k|≤n

vk

≤ X

m+1≤|k|≤n

kv_kk = t_n− t_m < .

This shows that (sn) forms a Cauchy sequence in (V, k · k). Since (V, k · k) is a Banach space,

(s_n) is convergent in (V, k · k). 

Corollary 1.1. (Comparison Test) Let (a_n)^∞_n=−∞ be a sequence of nonnegative real num- bers. Suppose that (vn)^∞_n=−∞ is a two sided sequence in a Banach space (V, k · k). Assume that

(1) kvnk ≤ a_n for any n ∈ Z (2) P∞

n=−∞a_nis convergent in R.

Then P∞

n=−∞vnis convergent in (V, k · k).

Proof. Let u_n=P

|k|≤na_k for n ≥ 0. Then (u_n) is convergent in R by assumption. Hence (u_n) is bounded above by some positive real number M. Let t_n =P

|k|≤nkv_nk for n ≥ 0.

For any n ≥ 0,

t_n+1 = t_n+ kv_n+1k + kv_−(n+1)k ≥ t_n.

We see that (tn) is nondecreasing. Since tn≤ u_nand un≤ M for any n ≥ 0, t_n≤ M. Hence (t_n) is nondecreasing and bounded above by M. By the monotone sequence property, (t_n) is convergent in R. Since (V, k · k) is a Banach space, by (1.1), P∞

n=−∞vn is convergent in

(V, k · k). 

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2. Application

Let 1 ≤ p < ∞. Let l^p(Z) be the space of all two sided complex sequences (an)^∞_n=−∞ such thatP∞

n=−∞|a_n|^p< ∞. On l^p(Z), we set kak_l^p=

∞

X

n=−∞

|a_n|^p

!1/p

.

Proposition 2.1. l^p(Z) forms a Banach space over C with respect to k · kl^p. When p = 2, we consider the inner product

ha, bi_l2 =

∞

X

n=−∞

anbn. Then kak_l2 =pha, ai_l2 for any a ∈ l²(Z).

Lemma 2.1. l¹(Z) is a subspace of l²(Z).

Proof. Let a = (a_n)^∞_n=−∞ be an element of l¹(Z), i.e. P∞

n=−∞|a_n| is convergent. Let s_n=P

|k|≤n|a_k| for n ≥ 0. Observe that for any n ≥ 0, s_n+1= s_n+ |a_n| + |a_(−n)|.

Since (s_n) is convergent,

n→∞lim(s_n+1− s_n) = lim

n→∞s_n+1− lim

n→∞s_n= 0.

This shows that lim_n→∞(|a_n| + |a_−n|) = 0. Choose  = 1. We can find N ∈ N so that

|a_n| + |a_−n| < 1 for any n ≥ N .

Therefore |a_n| < 1 and |a_−n| < 1 for n ≥ N. This shows that |a_n|²< |a_n| and |a_−n|² ≤ |a_−n| for any n ≥ N. By comparison test and the convergence of P∞

n=−∞|a_n|, P

|n|≥N|a_n|² is convergent and hence

∞

X

n=−∞

|a_n|² = X

|n|≤N −1

|a_n|²+ X

|n|≥N

|a_n|²

is convergent. This implies that a ∈ l²(Z). 

For each a = (an)^∞_n=−∞, we set

f (x) =

∞

X

n=−∞

ane^inx.

Then f converges in (C([0, 2π], C), k · k∞) by Proposition 1.1. Let sn(x) be the function s_n(x) = X

|k|≤n

a_ke^ikx, x ∈ [0, 2π].

For x ∈ [0, 2π],

|s_n(x)| ≤ X

|k|≤n

|a_k| ≤

∞

X

n=−∞

|a_n| = kak_l1.

Therefore ks_nk_∞ ≤ kak_l1. Since (s_n) is convergent to f in (C([0, 2π], C), k · k∞), kf k∞ ≤ kak_l1.

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We define a map T : l¹(Z) → C([0, 2π], C) T (a)(x) =

∞

X

n=−∞

a_ne^inx, x ∈ [0, 2π].

Then T is a linear map such that kT (a)k∞≤ kak_l1. In other words, T is a bounded linear map. On C([0, 2π], C), we consider the inner product

hf, gi_L2 = 1 2π

Z 2π 0

f (x)g(x)dx.

For any f ∈ C([0, 2π], C), we can prove

kf k_L2 ≤ kf k∞.

Hence if (fn) is a sequence convergent to f in (C([0, 2π], C), k·k∞), then (fn) is convergent to (C([0, 2π], C), k · kL²). This implies that ifP∞

n=−∞a_ne^inx is convergent in (C([0, 2π], C), k · k∞), it is convergent in (C([0, 2π], C), k · kL²). If f (x) = P∞

n=−∞ane^inx is convergent in (C([0, 2π], C), k · k∞), by the uniform convergence of f,

hf, e_mi = 1 2π

Z 2π 0

f (x)e^−imxdx =

∞

X

n=−∞

a_n· 1 2π

Z 2π 0

e^i(n−m)xdx = a_m. Here e_n(x) = e^inx for any n ∈ Z. Furthermore,

hf, s_mi_L2 = X

|k|≤m

akhf, e_ki = X

|k|≤m

|a_k|². This shows that

1 2π

Z 2π 0

f (x)s_m(x)dx = X

|k|≤m

|a_k|².

By taking m → ∞, and the uniform convergence of (sm) to f, we find kf k²_L2 = lim

m→∞

1 2π

Z 2π 0

f (x)s_m(x)dx = lim

m→∞

X

|k|≤m

|a_k|²=

∞

X

n=−∞

|a_n|².

In fact, let a, b ∈ l¹(Z) and let f (x) =P∞

n=−∞ane^inx and g(x) =P∞

n=−∞bne^inx, then hf, gi_L2 = ha, bi_l².

This can be proved by the uniform convergent of f (x) and g(x).