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)

\begin{align} \quad X / M = \{ x + M : x \in X \} \end{align}

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

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

(2)

\begin{align} \quad (x + M) + (y + M) = (x + y) + M \end{align}

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

(3)

\begin{align} \quad \alpha (x + M) = (\alpha x) + M \end{align}

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:

(4)

\begin{align} \quad \| \alpha (x + M) \| = \| (\alpha x) + M \| = \inf \{ \| \alpha x + m \|_X : m \in M \} = \inf \left \{ |\alpha| \left \| x + \frac{1}{\alpha} m \right \|_X : m \in M \right \} = |\alpha| \inf \{ \| x + m \|_X : m \in M \} = |\alpha| \| x + M \| \end{align}

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

(5)

\begin{align} \quad \quad \| (x + M) + (y + M) \| = \| (x + y) + M \| = \inf \{ \| (x + y) + m \|_X : m \in M \} &= \inf \{ \| (x + y) + 2m \|_X : m \in M \} \\ &\leq \inf \{ \| x + m \|_X + \| y + m \|_X : m \in M \} \\ & \leq \inf \{ \| x + m \|_X : m \in M \} + \inf \{ \| y + m \|_X : m \in M \| \} \\ & \leq \| x + M \| + \| y + M \| \end{align}

• 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:

(6)

\begin{align} \quad \| Q(x) \| = \| x + M \| = \inf \{ \| x + m \|_X : m \in M \} \leq \| x \|_X \end{align}

• 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:

(7)

\begin{align} \quad \| Q(x_0) - Q(x) \| = \| (x_0 + M) - (x + M) \| = \| (x_0 - x) + M \| = \| Q(x_0 - x) \| \leq \| x_0 - x \|_X < \delta = \epsilon \end{align}

• 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:

(8)

\begin{align} \quad \| (x_{n_k} + M) - (x_{n_{k+1}} + M) \| < \frac{1}{2^k} \end{align}

• 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:

(9)

\begin{align} \quad \| (x_{n_1} + y_1) - (x_{n_2} + y_2) \| \leq \| (x_{n_1} - x_{n_2}) + M \| + \frac{1}{2} < \frac{1}{2} + \frac{1}{2} = 1 \end{align}

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

(10)

\begin{align} \quad \| (x_{n_2} + y_2) - (x_{n_3} + y_3) \| \leq \| (x_{n_2} - x_{n_3}) + M \| + \frac{1}{4} < \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \end{align}

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

(11)

\begin{align} \quad \| (x_{n_k} + y_k) - (x_{n_{k+1}} + y_{k+1}) \| < \frac{1}{2^{k-1}} \end{align}

• 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 $X/M$. $\blacksquare$