The Difference Space of a Linear Space X Modulo a Linear Subspace L

# The Difference Space of a Linear Space X Modulo a Linear Subspace L

Definition: Let $X$ be a linear space and let $L \subseteq X$ be a linear subspace. Define an equivalence relation $\sim$ on $X$ for all $x, y \in X$ by $x \sim y$ if and only if $x - y \in L$. For each $x \in X$, the $L$-Coset of $x'$ denoted by $x'$ is the equivalence class containing $x$, that is, $x' = \{ y \in X : x - y \in L \}$. The Set of all $L$-Cosets of $X$ is denoted by $X - L$. |

Definition: Let $X$ be a linear space and let $L \subseteq X$ be a linear subspace. The Difference Space of $X$ Modulo $L$ is the space $X - L$ with the operation of addition defined for all $z, w \in X - L$ by $z + w = (x + y)'$ (where $x \in z'$ and $y \in w'$), and the operation of scalar multiplication defined for all $z \in X - L$ and all $\alpha \in \mathbf{F}$ by $\alpha z = (\alpha x)'$ (Where $x \in z'$). |

Let $X$ be a normed linear space and let $L \subseteq X$ be a linear subspace. If $L$ is closed then $X - L$ is a normed linear space with norm defined for all $z \in X - L$ by:

(1)\begin{align} \quad \| z \| = \inf \left \{ \| x \| : x \in z \right \} \end{align}

We can then define a linear operator from $X$ to $X - L$.

Definition: Let $X$ be a normed linear space and let $L \subseteq X$ be a closed linear subspace of $X$. The Canonical Map is the map $q : X \to X - L$ defined for all $x \in X$ by $q(x) = x'$. |

Proposition 1: Let $X$ be a normed linear space and let $L \subseteq X$ be a closed linear subspace. Let $q : X \to X - L$ be the canonical map. Then:a) If $L$ is a proper subspace of $X$ then $\| q \| = 1$.b) If $L = X$ then $\| q \| = 0$. |

**Proof of a)**Suppose that $L$ is a proper subspace of $X$. For each $x \in X$ we have that:

\begin{align} \quad \| q(x) \| = \| x' \| = \inf \{ \| y \| : y \in x' \} \leq \| x \| \end{align}

- (Where the inequality comes from the fact that $x \in x'$). So $\| q \| \leq 1$.

- On the other hand, since $L$ is a proper subspace of $X$, we can take $z \in X \setminus L$.

- Since $L$ is closed we have that $\| q(z) \| > 0$. Let $\epsilon > 0$ be such that $\epsilon < 1$. Then:

\begin{align} \quad \frac{\| q(z) \|}{1 - \epsilon} > \| q(z) \| = \inf \{ \| y \| : y \in z' \} \end{align}

- By the definition of infimum, there must exist a point $y \in z'$ such that:

\begin{align} \quad \frac{\| q(z) \|}{1 - \epsilon} &> \| y \| \\ \quad \| q(z) \| &> \| y \| (1 - \epsilon) \end{align}

- But since $y \in z'$ we have that $\| q(y) \| = \| q(z) \|$, and from above we have that:

\begin{align} \quad \| q(y) \| & > \| y \| (1 - \epsilon) \end{align}

- Since $z - y \in L$ we see that $y \neq 0$ otherwise $z$ would be in $L$ (which it cannot be since $z \in X - L$). Thus $\| y \| \neq 0$. Since for all $0 < \epsilon < 1$ there exists such a $y$ which makes the above inequality hold, we conclude that $\| q \| \geq 1$.

- Thus $\| q \| = 1$. $\blacksquare$

**Proof of b)**Suppose that $L = X$. Let $x \in X$. Then for every $y \in X$ we have that $x - y \in X = L$. So the only $L$-coset is $0'$.

- If $q : X \to X - L$ is the canonical map then $q(x) = 0$ for every $x \in X$ and so clearly $\| q \| = 0$. $\blacksquare$