Subspace of Normed Vector Space with Induced Norm forms Normed Vector Space

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Theorem

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.

Let $Y \subseteq X$ be a vector subspace.

Let $\norm {\, \cdot \,}_Y$ be the induced norm on $Y$.


Then $\struct {Y, \norm {\, \cdot \,}_Y}$ is a normed vector space.


Proof

Positive definiteness

By definition of induced norm:

$\forall y \in Y : \norm {y}_Y = \norm {y}_X > 0$

Suppose $y \in Y: \norm y_Y = 0$.

Since $\norm {\, \cdot \,}_Y$ is an induced norm in $\struct {X, \norm {\, \cdot \,}_X}$:

$\norm y_X = 0$

Therefore:

$y = \mathbf 0 \in X$

By definition of a vector subspace, $Y$ has the additive identity:

$\mathbf 0 \in Y$

Thus:

$y = \mathbf 0 \in Y \subseteq X$


Positive homogeneity

Let $y \in Y$.

Let $\alpha \in \R$.

Then we have that:

\(\ds \norm {\alpha \cdot y}_Y\) \(=\) \(\ds \norm {\alpha \cdot y}_X\) Definition of Induced Norm
\(\ds \) \(=\) \(\ds \size \alpha \norm y_X\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \size \alpha \norm y_Y\) Definition of Induced Norm; $y \in Y$


Triangle inequality

Let $y_1, y_2 \in Y$.

By closure axiom of vector space:

$y_1 + y_2 \in Y$

Furthermore:

\(\ds \norm {y_1 + y_2}_Y\) \(=\) \(\ds \norm {y_1 + y_2}_X\) Definition of Induced Norm
\(\ds \) \(\le\) \(\ds \norm {y_1}_X + \norm {y_2}_X\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \norm {y_1}_Y + \norm {y_2}_Y\) Definition of Induced Norm; $y_1, y_2 \in Y$

$\blacksquare$


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