# Quotient Normed Linear Spaces

Let $(X, \| \cdot \|_X)$ be a normed linear space and let $M$ be a linear subspace of $X$. We ultimately want to turn $X / M$ into a normed linear space where:

(1)Where $x + M = \{ x + m : m \in M \}$.

We define addition of $x + M, y + M \in X/M$ by:

(2)And for all $\alpha \in \mathbb{R}$ we define scalar multiplication by:

(3)The following proposition tells us that we can always define a particular seminorm on $X / M$.

Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $M$ be a linear subspace of $X$. Define $\| \cdot \| : X / M \to [0, \infty)$ for all $x + M \in X/M$ by $\| x + M \| = \inf \{ \| x + m \|_X : m \in M \}$. Then $\| \cdot \|$ is a seminorm on $X / M$. |

**Proof:**There are two things to show.

- First let $\alpha \in \mathbb{R}$ and let $x + M \in X/M$. If $\alpha = 0$ then $\| \alpha (x + M) \| = \| 0 + M \| = \inf \{ \| m \|_X : m \in M \} = 0$ since $M$ is a subspace of $X$ and so $0 \in M$. If $\alpha \neq 0$ then since $\frac{1}{\alpha}m \in M$ we have that:

- Now let $x + M, y + M \in X / M$. Then:

- So $\| \cdot \|$ is a seminorm on $X / M$. $\blacksquare$

Note that in general, $\| \cdot \|$ need not be a norm on $X/M$, that is, it may be that $\| x + M \| = 0$ does not imply that $x + M = 0 + M$. If $M$ is a closed subspace of $X$ then it is.

Proposition 2: If $(X, \| \cdot \|_X)$ is a normed linear space and let $M$ be a closed linear subspace of $X$. Then $\| \cdot \| : X/M \to [0, \infty)$ defined for all $x + M \in X/M$ by $\| x + M \| = \inf \{ \| x + m \|_X : m \in M \}$ is a norm on $X/M$. |

**Proof:**Proposition 1 establishes that $\| \cdot \|$ is a seminorm on $X/M$. So all that remains to show is that $\| x + M \| = 0$ if and only if $x + M = 0 + M$.

- First suppose that $\| x + M \| = 0$. Then $\inf \{ \| x + m \|_X : m \in M \} = 0$. Observe that $\| x + m \|_X = 0$ if and only if $x + m = 0$, that is, $m = -x$. So there must exist a sequence $(m_n) \subset M$ such that $(m_n)$ converges to $-x$. Since $M$ is closed, $-x \in M$. Since $M$ is a subspace of $X$ we have that $x \in M$. But then $x + M = 0 + M$.

- Conversely, suppose that $x + M = 0 + M$. Then $\| x + M \| = \| 0 + M \| = \inf \{ \| m \|_X : m \in M \}$. Since $M$ is a subspace of $X$ we have that $0 \in M$ and so $\| 0 + M \| = 0$. $\blacksquare$

If $M$ is not closed, then $M$ does not contain all of its accumulation points. Let $t$ be an accumulation point of $M$. Note that $t \neq 0$ since $0 \in M$ by the definition of $M$ being a subspace. Let $(m_n)$ be a sequence in $M$ which converges to $t$. Then $\inf \{ \| t - m \|_X : m \in M \} = 0$. So $\| t + M \| = \| 0 + M \|$. But $t + M \neq 0 + M = M$ since $t \not \in M$.

Definition: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $M$ be a linear subspace of $X$. The Quotient Map from $X$ to $X/M$ is defined to be the map $Q : X \to X/M$ defined for all $x \in X$ by $Q(x) = x + M$. |

*Sometimes the map $Q$ defined above is called the Natural Map from $X$ to $X/M$.*

Proposition 3: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $M$ be a closed linear subspace of $X$. Then $\| Q(x) \| \leq \| x \|$ for all $x \in X$, and $Q$ is continuous on $X$. |

**Proof:**Let $x \in X$. Then:

- Since $0 \in M$.

- For continuity, let $x_0 \in X$ and let $\epsilon > 0$ be given. Suppose that $\delta = \epsilon$. If $\| x_0 - x \|_X < \delta = \epsilon$ then from above we have that:

- So $Q$ is continuous at $x_0$ and hence $Q$ is continuous on all of $X$. $\blacksquare$

Theorem 4: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $M$ be a closed linear subspace of $X$. If $X$ is a Banach space then $X/M$ is a Banach space. |

**Proof:**Let $(x_n + M)$ be a Cauchy sequence in $X/M$. Then there must exist a subsequence $(x_{n_k}+ M)$ of $(x_n + M)$ such that:

- Set $y_1 = 0$. Consider the value $\| (x_{n_1} - x_{n_2}) + M \| + \frac{1}{2}$. Note that $\| (x_{n_1} - x_{n_2}) + M \| \leq \| (x_{n_1} - x_{n_2}) + M \| + \frac{1}{2}$ clearly, and so there exists a $y_2 \in M$ for which:

- Now consider the value $\| (x_{n_2} - x_{n_2}) + M \| + \frac{1}{4}$. We can choose a $y_3 \in M$ for which:

- We continue on in this manner to obtain a sequence $(y_k)$ in $M$ for which:

- Therefore $(x_{n_k} + y_k)$ is a Cauchy sequence in $X$, and since $X$ is a Banach space there exists an $x \in X$ such that $(x_{n_k} + y_k)$ converges to $x$. Since $Q$ is continuous by the previous proposition, we have that $Q(x_{n_k} + y_k)$ converges to $Q(x)$, that is, $(x_{n_k} + y_k) + M$ converges to $x + M$.

- Since $y_k \in M$ for each $k$ we have that $(x_{n_k} + M)$ converges to $x_0 + M \in X/M$. So every Cauchy sequence in $X/M$ converges in $X/M$. $\blacksquare$