15.5 Sequence Spaces

Definition 15.5.1 ($c_{0}$-Direct Sum).label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For any $x \in \prod_{i \in I}X_{i}$, $x$ vanishes at infinity if for each $\eps > 0$, $\bracs{i \in I| \norm{x_i}_{X_i} \ge \eps}$ is finite. The space

\[[c_{0}(I); X_{i}] = \bracs{x \in \prod_{i \in I}X_i \bigg | x \text{ vanishes at infinity}}\]

equipped with the uniform norm

\[\norm{x}_{[c_0(I); X_i]}= \sup_{i \in I}\norm{x_i}_{X_i}\]

is the $c_{0}$-direct sum of $\seqi{X}$.

Definition 15.5.2 ($l^{p}$-Direct Sum).label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the $l^{p}$-direct sum of $\seqi{X}$ is the space

\[[l^{p}(I); X_{i}] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sum_{i \in I}\norm{x_i}_{X_i}^p < \infty}\]

equipped with the norm

\[\norm{x}_{[l^p(I); X_i]}= \braks{\sum_{i \in I}\norm{x_i}_{X_i}^{p}}^{1/p}\]

Definition 15.5.3 ($l^{\infty}$-Direct Product).label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the $l^{\infty}$-direct product of $\seqi{X}$ is the space

\[[l^{\infty}(I); X_{i}] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sup_{i \in I}\norm{x_i}_{X_i} < \infty}\]

equipped with the norm

\[\norm{x}_{[l^\infty(I); X_i]}= \sup_{i \in I}\norm{x_i}_{X_i}\]

Proposition 15.5.4.label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty]$.

  1. (1)

    (Hölder’s Inequality) Let $q \in [1, \infty]$ be the Hölder conjugate of $p$, $\seqi{Y}, \seqi{Z}$ be normed spaces, and $\seqi{\lambda}$ such that for each $i \in I$, $\lambda \in L^{2}(X_{i}, Y_{i}; Z)$.

    For each $x \in [l^{p}(I); X_{i}]$ and $y \in [l^{q}(I); Y_{i}]$, let $\lambda(x, y)_{i} = \lambda_{i}(x_{i}, y_{i})$, then

    \[\norm{\lambda(x, y)}_{[l^1(I); Z_i]}\le \norm{x}_{[l^p(I); X_i]}\cdot \norm{y}_{[l^q(I); X_i]}\cdot \sup_{i \in I}\norm{\lambda_i}_{L^2(X_i, Y_i; Z_i)}\]

  2. (2)

    (Minkowski’s Inequality) For each $x, y \in [l^{p}(I); X_{i}]$,

    \[\norm{x + y}_{[l^p(I); X_i]}\le \norm{x}_{[l^p(I); X_i]}+ \norm{y}_{[l^p(I); X_i]}\]

  3. (3)

    (Markov’s Inequality) If $p < \infty$, then for each $\alpha > 0$ and $x \in [l^{p}(I); X_{i}]$,

    \[|\bracsn{i \in I|\ \norm{x}_{X_i} \ge \alpha}| \le \frac{1}{\alpha^{p}}\norm{f}_{[l^p(I); X_i]}^{p}\]

    In particular, $\bracs{i \in I|x_i \ne 0}$ is countable.

  4. (4)

    For any $q \in [p, \infty]$, $[l^{p}(I); X_{i}] \subset [l^{q}(I); X_{i}]$, where for any $x \in [l^{p}(I); X_{i}]$, $\norm{x}_{[l^q(I); X_i]}\le \norm{x}_{[l^p(I); X_i]}$.

  5. (5)

    If $X_{i}$ is a Banach space for all $i \in I$, then so is $[l^{p}(I); X_{i}]$.

Proof. (1), (2), (3): By the classical Hölder’s inequality, Minkowski’s inequality, and Markov’s inequality.

(4): For each $i \in I$, $\norm{x_i}_{X_i}\le \norm{x}_{[l^p(I); X_i]}$, so the result holds when $q = \infty$.

If $q < \infty$, then by Proposition 15.1.12, there exists $\lambda \in [p, q]$ such that

\[\norm{x}_{[l^p(I); X_i]}\le \norm{x}_{[l^p(I); X_i]}^{\lambda}\norm{x}_{[l^\infty(I); X_i]}^{1 - \lambda}\le \norm{x}_{[l^p(I); X_i]}\]

$\square$

Theorem 15.5.5.label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For each $y \in [l^{1}(I); X_{i}^{*}]$, let

\[\phi_{y}: [c_{0}(I); X_{i}] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}\]

then the mapping

\[[l^{1}(I); X_{i}^{*}] \to [c_{0}(I); X_{i}]^{*} \quad y \mapsto \phi_{y}\]

is an isometric isomorphism.

Proof. By Hölder’s Inequality, for each $y \in [l^{1}(I); X_{i}^{*}]$, $\norm{\phi_y}_{[c_0(I); X_i]^*}\le \norm{y}_{[l^1(I); X_i^*]}$.

Let $\phi \in [c_{0}(I); X_{i}]^{*}$, then there exists $y \in [l^{\infty}(I); X_{i}^{*}]$ such that for each $i \in I$ and $x_{i} \in X_{i}$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[c_0(I); X_i]}= \dpn{x_i, y_i}{X_i}$.

Let $J \subset I$ be finite and $\alpha \in (0, 1)$, then there exists $x \in [c_{0}(I); X_{i}]$ such that

  1. (1)

    $\{i \in I|x_{i} \ne 0\} \subset J$.

  2. (2)

    $\norm{x}_{[c_0(I); X_i]}\le 1$.

  3. (3)

    For each $j \in J$, $\dpn{x_j, y_j}{X_j}\ge \alpha\norm{y_j}_{X_j^*}$.

Thus

\[\alpha\sum_{j \in J}\norm{y_j}_{X_j^*}\le \sum_{j \in J}\dpn{x_j, y_j}{X_j}= \dpn{x, \phi}{[c_0(I); X_i]}\le \norm{\phi}_{[c_0(I); X_i]^*}\]

As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^{1}(I); X_{i}^{*}]$ with $\norm{y}_{[l^1(I); X_i^*]}\le \norm{\phi}_{[c_0(I); X_i]^*}$. By the Dominated Convergence Theorem, $\dpn{x, \phi_y}{[c_0(I); X_i]}= \dpn{x, \phi}{[c_0(I); X_i]}$ for all $x \in [c_{0}(I); X_{i}]$. Hence the map is an isometric isomorphism.$\square$

Theorem 15.5.6.label Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$ and $q \in (1, \infty]$ be Hölder conjugates. For each $y \in [l^{q}(I); X_{i}^{*}]$, let

\[\phi_{y}: [l^{p}(I); X_{i}] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}\]

then the mapping

\[[l^{q}(I); X_{i}^{*}] \to [l^{p}(I); X_{i}]^{*} \quad y \mapsto \phi_{y}\]

is an isometric isomorphism.

Proof. Let $\phi \in [l^{p}(I); X_{i}]^{*}$, then there exists $y \in [l^{\infty}(I); X_{i}^{*}]$ such that for each $i \in I$ and $x_{i} \in X_{i}$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]}= \dpn{x_i, y_i}{X_i}$.

Since the $q = \infty$ case has been ruled out, assume that $q \in (1, \infty)$. For each $\alpha \in (0, 1)$, there exists $x \in [l^{\infty}(I); X_{i}]$ with $\norm{x_i}_{X_i}\le 1$ and $\dpn{x_i, y_i}{X_i}\ge \alpha \norm{y_i}_{X_i^*}$. For each $J \subset I$ finite and $i \in I$, let $F_{J}(i) = \one_{J}(i) \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by Lemma 15.1.4,

\[\norm{F_J x}_{[l^p(I); X_i]}^{p} \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}\]

so

\begin{align*}\alpha \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}&\le \sum_{i \in I}F_{J}(i)\dpn{x_i, y_i}{X_i}= \dpn{F_J x, \phi}{[l^p(I); X_i]}\\&\le \norm{\phi}_{[l^p(I); X_i]^*}\cdot \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^{q}}^{1/p}\\ \alpha \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^q}^{1/q}&\le \norm{\phi}_{[l^p(I); X_i]^*}\end{align*}

As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^{q}(I); X_{i}^{*}]$ with $\norm{y}_{[l^q(I); X_i^*]}= \norm{\phi}_{[l^p(I); X_i]^*}$.$\square$

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