On an exam today I used that if $X=\mathcal{C}[a,b]$ and $Y=\{f\in X : f(a)=f(b)\}$, then the projection $\pi: X\rightarrow X/Y$ has the property $\ker(\pi)=Y$. This led me to the following:
Suppose $(V, ||\cdot ||)$ is an infinite dimensional normed space and $W$ is a subspace of $V$ with $\pi:V\rightarrow V/W$ the canonical projection. Is there a canonical choice of norm on the space $V/W$ (say where $||\pi||=1$)? If there is such a norm, does $\ker(\pi)=W$?
My speculation is, ignoring some details of the quotient structure, if there is an accumulation point, call it $p$, of a sequence in $Y$ with $p\not\in Y$, then in the quotient space this would provide a sequence identically $0$, so $\pi(y)$ would necessarily be $0$.
I haven't really seen much about this type of problem before, thanks for the help.
In case that $W$ is closed, there is indeed a canonical choice. It is called the quotient norm and defined by \begin{equation*} \| [\hat v] \|_{V / W} := \inf_{v \in [\hat v]} \| v \|_V . \end{equation*} This also leads to $\|\pi\| = 1$ (in case $W \ne V$).
As already mentioned by matt biesecker, you always have $\operatorname{ker}(\pi) = W$, since this is independent of the norm.
This implies that if $W$ is not closed, you cannot define a norm such that $\pi$ is continuous. This would imply $W = \operatorname{ker}(\pi)$ is closed.