# 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 $C([a,b])$

-
To prove that the quotient (semi)norm is a norm, you'll need to use that $M$ is closed in $X$. – Keenan Kidwell Dec 12 '12 at 16:33
I know that if M is closed in X, than $||[x]||$ is a norm on X\M, but how can I show that? – Voyage Dec 12 '12 at 16:36
If $x+M=M$, i.e., $x\in M$, then it is clear that $\mathrm{dist}(x,M)=0$. If $x+M\neq M$, meaning $x\notin M$, then there is some $\epsilon>0$ such that the closed ball of radius $\epsilon$ centered around $x$ does not intersect $M$ (this is where I used that $M$ is closed because $x\notin M$ is equivalent to $x\notin\overline{M}$). This means that for $y\in M$, $\vert\vert x-y\vert\vert>\epsilon$. So $\mathrm{dist}(x,M)\geq\epsilon>0$. – Keenan Kidwell Dec 12 '12 at 17:46
So in the case that $x+M\not=M$ the quotient norm is not a norm, but if $x+M=M$ than it is? – Voyage Dec 12 '12 at 18:56
Dear @Voyage, No. The elements of $X/M$ are cosets $x+M$, for $x\in X$. If $M$ is closed (it isn't by Matt N.'s argument, although I originally assumed you were using the supremum norm, where $M$ is closed), then my argument above shows that $\vert\vert x+M\vert\vert=0$ if and only if $x+M=M$, which is to say that the kernel of the quotient semi-norm is zero, so it is an actual norm. – Keenan Kidwell Dec 12 '12 at 19:48

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.

-
Thank you, so the quotient norm is semi norm. I could show the triangle inequality, the homogenity but if $inf||x+m||=0$ it does not follow that $x+m=0$. $||x_n-(-x)||_{L^1}=\int_{0}^{1}|x_n(t)+x|dt->0$ How can this x_n now be constructed ? To 3) Actually $a\le\alpha\le\beta\le b$ and $\forall s\in[\alpha,\beta]$ – Voyage Dec 12 '12 at 17:11
@Voyage I edited my answer. As for 3): I don't know what you mean. It's also not clear what $a,b,\alpha, \beta$ are. I assume $a=0$ and $b=1$. For other values the proof remains the same. – Rudy the Reindeer Dec 12 '12 at 19:36