Direct Product of Banach Spaces is Banach Space

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Theorem

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.

Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.

Let $\norm {\, \cdot \,}_{X \times Y}$ be the direct product norm.

Suppose $X$ and $Y$ are Banach spaces.


Then $V$ is a Banach space.


Proof

Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a Cauchy sequence in $V$:

$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }_{X \times Y} < \epsilon$

We have that:

\(\ds \norm {x_n - x_m}_X\) \(\le\) \(\ds \map \max {\norm {x_n - x_m}_X, \norm {y_n - y_m}_Y }\)
\(\ds \) \(=\) \(\ds \norm {\tuple {x_n - x_m, y_n - y_m} }_{X \times Y}\) Definition of Direct Product Norm
\(\ds \) \(=\) \(\ds \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }_{X \times Y}\) induced component-wise operations
\(\ds \) \(<\) \(\ds \epsilon\)

Hence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy sequence in $X$.

By assumption, $X$ is a Banach space.

Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.

By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in Y$.

So, by Convergence in Direct Product Norm, we have:

$\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ converges to $\tuple {x, y}$.

By definition, $V$ is a Banach space.

$\blacksquare$


Sources

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