§ Quotient spaces of Banach space

• We will see why it is important for a subspace $M$ of a banach space $X$to be closed for $X/M$ to be banach.
• The algberaic properties of $+$ and $\cdot$ will go through for any subspace $M$ since they in no way depend on norm.
• The norm on $X/M$ will correctly interact with rescaling and triangle inequality also for any subspace $M$.
• However, to show that the norm is non-degenerate ( $||x|| = 0$ iff $x = 0$) needs $M$ to be closed.

§ Norm on $X/M$

• We define the norm on $X/M$ as $||\overline{x}|| \equiv \inf_{m \in M} ||x + m||$. This is abbreviated to $||x + M||$.

§ Lemma: Norm on $X/M$ interacts correctly with rescaling

• $||\alpha \overline{x}|| = \inf_{m \in M} ||\alpha x + m||$.
• But we can replace $m \mapsto \alpha m$, giving $\inf_{m \in M} || \alpha x + \alpha m||$, which equals $\inf_{m \in M} \alpha ||x + M|| = \alpha || \overline{x}||$.
• Thus, scalar product correctly rescales with norm.

§ Lemma: Norm on $X/M$ obeys triangle ineq

• The LHS is $||\overline{x} + \overline{y}|| = \inf_{m \in M} ||x + y + m||$.
• The RHS is $||\overline{x}|| + ||\overline{y}|| = \inf{k \in M} || x + k|| + \inf_{l \in M} ||y + l||$.
• We need to somehow "split" the $m$ in the LHS into $k$ and $l$.
• We do this sequentually. There must be a sequence of elements $k[i]$ such that $||\overline{x}|| \leq ||x + k[i]||$ such that $||x + k[i]|| \to ||\overline{x}||$.
• Similarly, there must be a sequence of elements $l[i]$ such that $||\overline{y}|| \leq ||y + l[i]||$ such that $||y + l[i]|| \to ||\overline{y}||$.
• Now, we see that $||overline{x} + \overline y|| \leq ||x + y + k[i] + l[i]||$.
• By triangle inequality, this is going to be $||\overline{x} + \overline{y}|| \leq ||x + k[i]|| + ||y + l[i]||$.
• Since this holds pointwise, it also holds in the limit, proving triangle inequality..

§ Theorem: proving that norm of zero is zero

• It is clear that $|| \overline 0|| = \inf_{m \in M} || 0 + m|| = || 0 + 0 || = 0$.

§ Theorem: proving that norm is nondegenerate.

• Suppose $||\overline{x}|| = 0$. We want to show that $\overline{x} = 0$, or $x \in M$.
• This means that $\inf_{m \in M} ||x + m|| = 0$.
• Thus there are a sequence of elements $m[i]$ such that $||x + m[i]|| \to 0$.
• This implies that $x + m[i] \rightarrow 0$, since this is happening using the norm of the underlying space.
• This means that $m[i] \to -x$.
• Now, we need to use the fact that $M$ is closed, to say that $-x \in M$, to get that $x \in M$.
• This gives us that $\overline{x} = 0$.