Closed Subspace of Banach Space forms Banach Space

Theorem
Let $\struct {X, \norm \cdot_X}$ be a Banach space.

Let $Y$ be a closed linear subspace of $X$.

Let $\norm \cdot_Y$ be the restriction of $\norm \cdot_X$ to $Y$.

Then $\struct {Y, \norm \cdot_Y}$ is a Banach space.

Proof
From Restriction of Norm on Vector Space to Subspace is Norm, $\struct {Y, \norm \cdot_Y}$ is a normed vector space.

We now show that $\struct {Y, \norm \cdot_Y}$ is complete.

Let $\sequence {y_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {Y, \norm \cdot_Y}$.

Then, for each $\epsilon > 0$ there exists $N \in \N$ such that:


 * $\norm {y_n - y_m}_Y < \epsilon$

for all $n, m \ge N$.

Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have:


 * $\norm {y_n - y_m}_Y = \norm {y_n - y_m}_X$

so:


 * $\norm {y_n - y_m}_X < \epsilon$

So $\sequence {y_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm \cdot_X}$.

Since $\struct {X, \norm \cdot_X}$ is Banach, we have that $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in X$.

Since $Y$ is closed in $\struct {X, \norm \cdot_X}$, we have that $Y$ contains all its limit points.

So $y \in Y$.

From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we therefore have:


 * $\norm {y_n - y}_X \to 0$

with $y_n - y \in Y$ for each $n \in \N$.

Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have:


 * $\norm {y_n - y}_Y \to 0$

So $\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $\struct {Y, \norm \cdot_Y}$.

Since $\sequence {y_n}_{n \mathop \in \N}$ was arbitrary, we have that all Cauchy sequence in $\struct {Y, \norm \cdot_Y}$ converge.

Also see

 * Subspace of Complete Metric Space is Closed iff Complete