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; $y_1, y_2 \in Y$ |
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces