$X$ is the vector space $C[0,1]$ with the norm $\|f\|_1=\int_0^1|f(t)|dt$, and $M=\{f\in X:f(0)=0\}$, show that $M$ is not closed. Here is my question:
Let $X$ be the vector space $C[0,1]$ with the norm $\|f\|_1=\int_0^1|f(t)|dt$. Let
$$M=\{f\in X:f(0)=0\}$$
Show that $M$ is not closed. Show that the “quotient norm" $inf\{\|f-m\|_1:m\in M\}$ is not a norm on $X/M$.
Here is what I have:
For the closure, I am trying to find a function $f$ such that a sequence $\{f_n\}\in M$ which converges to $f$, but $f\notin M$. This would mean that for any $n$, $f_n(0)=0$ but $f(0)\neq 0$. I cannot seem to find such an $f_n$ and $f$.
As for the quotient norm not being a norm on $X/M$, I believe this follows from $M$ not being closed, as with quotient norms, $\|x-M\|=0$ if and only if $x\in M$.
So any suggestions on finding the $f$? Thanks.

As for the "quotient norm", have a look at mookid's counter-example and see if you can modify it to force a violation of one of the norm axioms. I'd aim to show that $$\| [f] \|_{\hbox{am I a norm?}} = 0  \ \ \ \Longrightarrow \ \ \  f = 0$$ 
is violated.
