Quotient norm on $X\backslash M$ I have $X=(C([0,1]),||.||_1)$ where $||f||_1=\int_{0}^{1}|f(t)|dt$ and $M=\{f\in C([a,b]): f(0)=0\}$. Now I have three questions:
1) Is the quotient norm a norm on the quotient space X\M ? 
What I tried was the following: The quotient space is defined by X\M=$\{[x]:x\in X\}$
There fore $||[x]||=dist(x,M)=\inf_{m\in M}||x-m||$ What do I have to do now?
2) What does the kernel of the quotient norm look like?
3) Can I follow from the points 1) and 2) that if $X=C([a,b])$ and $M=\{f\in C([a,b]): f(s)=0 \forall s\in(a,b)\}$ X\M is isometrically isomoprh to $C([a,b])$

1) The quotient norm is defined as $\|f + M \|_{X/M} = \inf_{g \in M} \|f+g\|$. This defines a norm if it satisfies the properties of a norm. In particular, it must hold that $\|f + M \|_{X/M} = \inf_{g \in M} \|f+g\|= 0$ if and only if $f + M =0$. 
Now you ask yourself how you could violate the above property. It is violated if you can find a sequence in $M$ that converges to $-f$ in $\|\cdot\|_{L^1}$ for an $f$ that is outside $M$.
Can you find such a sequence? Edit In response to your comment: Fix $\delta > 0$. Take $f_n$ to be the function that is $0$ on $[0,1/n - \delta]$, linear on $[1/n - \delta, 1/2 - \delta]$ and $1$ otherwise. Then these $f_n$ are Cauchy in $L_1$, continuous, zero at zero but the limit function is non-zero at zero. 
2) Well, two cases. Either the quotient norm is a norm and its kernel is trivial or it is a semi-norm and the kernel is non-trivial. You will know after answering 1).
3) I am unsure about your last edit and point 3): continuous functions that are zero on all of $(a,b)$ are also zero on $[a,b]$. Then $M = \{0\}$. Then of course $X$ is isometrically isomorphic to itself: via the identity map. 
I hope I didn't misunderstand your question. 
