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Here is my question:

Let $X$ be a Banach space $C[0,1]$ with the supremum norm. Let $M=\{f\in X: f(0)=0\}$. Show that $M$ is closed. Find an explicit formula for the quotient norm $\|[f]\|$ for $[f]\in X/M$. Find an isometric isomorphism from $\mathbb{R}$ to $X/M$.

Here is what I have:

$M$ is closed: Let $f_n\in M$ such that $f_n\to f$. Let us assume that $M$ is open, that is to say that $f(0)=y\neq 0$. Then, given $\epsilon >0$ we know there exists some $N>$ such that for a fixed $x\in [0,1]$, $\|f_n-f\|<\epsilon$ for all $n\geq N$. Set $\epsilon < y$ and choose $x=0$. Then:

$$\|f_n(0)-f(0)\|=\|0-y\|=sup\{|-y|\}=y\gt\epsilon$$

So we have a contradiction, therefore $f(0)=0$ and $M$ is closed.

Explicit formula for the quotient norm:

$$\|[f]\|=\|f+m\|=inf\{\|f+m\|_\infty:m\in M\}=inf\{sup\{|f+m|\}:m\in M\}$$

Isomorphism:

This one I am having some trouble with.

Indrajit
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user3784030
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    For the isomorphism, try proving that the inverse of the evaluation at zero map $X/M\rightarrow\mathbb{R}$ works. Btw, the quotient norm is $\Vert [f]\Vert=\vert f(0)\vert$. – J. David Taylor Oct 31 '14 at 01:34

1 Answers1

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Continuing your solution: First I will show that $\|[f]\|=|f(0)|$. Notice that for a fixed $f\in C[0,1]$ and for any $m\in M$, we have $\|f+m\|_{\infty}=\sup_{x\in[0,1]}|f(x)+m(x)|\geq |f(0)+m(0)|=|f(0)|$. Therefore $\|[f]\|=\inf\{\|f+m\|_{\infty}:m\in M\}\geq |f(0)|$. Conversely, define the function $m(x)=f(0)-f(x)$. Clearly $m\in M$, and $\|f+m\|_{\infty}=|f(0)|$. Therefore $\|[f]\|=|f(0)|$.

Now define a function $F:X\to\mathbb{R}$ as $F(f)=f(0)$. Obviously $\ker(F)=M$. So $X/M$ is isomorphic to $\mathbb{R}$. This isomorphism is also an isometry, which follows from the fact that $\|[f]\|=|f(0)|$.

Indrajit
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