If $X=C[0,1]$ and $X_0=\{f\in C[0,1]|f(0)=0$, then how to show that $X/X_0$ is isometrically isomorphic to $\mathbb C$?

Question

Of course, $X/X_0$ is the quotient Banach space with usual norm. I think it's true that the map should be one that takes a continuous function to its value at 0. But can someone give a precise argument showing that the map will be an isometric isomorphism? Thanks in advance.

Answer 1

Let's first notice that $X_0$ is a closed subspace of $X$. Hence we can define the quotient norm by $$\Vert \overline{f} \Vert = \inf_{g \in X_0} \Vert f- g \Vert$$... with an obvious abuse of notation.

The application $$ \begin{array}{l|rcl} \phi : & X/X_0 & \longrightarrow & \mathbb C\\ & \overline{f} & \longmapsto & f(0) \text{ for any } f \in \overline{f}\end{array}$$ is well defined as for any $\overline{f} \in X/X_0$ and $f_1,f_2 \in \overline{f}$ we have $f_1(0)-f_2(0)=0$.

If we prove that for $\overline{f} \in X/X_0$, we have $\Vert \overline{f} \Vert = \vert f(0) \vert=\vert \phi(\overline{f}) \vert $, we will have proven the required result, i.e. that $\phi$ is an isometry between $X/X_0$ and $\mathbb C$. The proof is not so complex using the sequence $(g_n) \in X_0^{\mathbb N}$ defined by $$ \begin{array}{l|rcl} g_n : & [0,1] & \longrightarrow & \mathbb C\\ & x & \longmapsto & f(x) \text{ for } x \in [\frac{1}{n},1] \\ & x & \longmapsto & nx f(\frac{1}{n}) \text{ elsewhere} \\ \end{array}$$ One can verify that $$\lim\limits_{ n\to \infty} \Vert f - g_n \Vert = \vert f(0) \vert$$ As $\Vert \overline{f} \Vert \ge \vert f(0) \vert$, we can conclude.