Restriction of Norm on Vector Space to Subspace is Norm

Theorem

Let $\Bbb F$ be a subfield of $\C$.

Let $X$ be a vector space over $\Bbb F$.

Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a norm on $X$.

Let $Y$ be a vector subspace of $Y$.

Then $\norm \cdot_Y$, the restriction of $\norm \cdot_X$ to $Y$ is a norm on $Y$.

Proof

Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.

We verify each of the axioms for a norm on a vector space.

Proof of $(\text N 1)$

Note that for $y \in Y$ we have $\norm y_Y = 0$ if and only if $\norm y_X = 0$.

Then by $(\text N 1)$ for $\norm \cdot_X$, we have $y = 0$.

$\Box$

Proof of $(\text N 2)$

Let $\lambda \in \Bbb F$ and $y \in Y$.

Then, we have:

\(\ds \norm {\lambda y}_Y\)

\(=\)

\(\ds \norm {\lambda y}_X\)

\(\ds \)

\(=\)

\(\ds \cmod \lambda \norm y_X\)

from $(\text N 2)$ for $\norm \cdot_X$

\(\ds \)

\(=\)

\(\ds \cmod \lambda \norm y_Y\)

$\Box$

Proof of $(\text N 3)$

Let $x, y \in Y$.

Then, we have:

\(\ds \norm {x + y}_Y\)

\(=\)

\(\ds \norm {x + y}_X\)

\(\ds \)

\(\le\)

\(\ds \norm x_X + \norm y_Y\)

from $(\text N 3)$ for $\norm \cdot_X$

\(\ds \)

\(=\)

\(\ds \norm x_Y + \norm y_Y\)

$\blacksquare$