$\ker f$ of linear functional in Banach space $f(x)=x(1)-x(0)$

Question

We have Banach space $C([0,1],\mathbb{R})$ and a linear functional $f:C([0,1],\mathbb{R})\rightarrow \mathbb{R}$ such as $f(x)=x(1)-x(0)$ for any $x \in C([0,1],\mathbb{R})$.

I have to find the $\ker f$, $\|f\|$, and $d(t,\ker f)$. So I wanted to start with $\ker f:=\{x\in X: f(x)=0\}$

$f(x) =0\iff x(1)-x(0)=0$.

So, $x(1)=x(0)$ and how we define that $\ker f$?

Answer 1

$\ker (f)= \{x\in C[0,1]: x(0)=x(1)\}$.

If $x \in \ker (f)$ then $\|t-x(t)\| \geq |0-x(0)|$ and $\|t-x(t)\| \geq |1-x(1)|$. Let $c=x(0)=x(1)$. Then $c \in [0,1]$ so either $|c| \geq \frac 12$ or $|1-c|\geq \frac 1 2$. Hence $\|t-x(t)\| \geq \frac 1 2 $. Since this is true for every $x \in \ker (f)$ it follows trhat $d(t, \ker (f)) \geq \frac 1 2$. Now let $x(t)=\frac 1 2 $ for all $t$. Then $x \in \ker (f)$ so $d(t, \ker f) \leq \|t-x(t)\|=\|t-\frac 12 \|=\frac 1 2$. We have proved that $d(t, \ker (f))=\frac 12 $.

Now $\|f\| \leq 2$ since $\|x\| \leq 1$ implies $|x(1)-x(0)| \leq 2$. If we take $x(2)=2(t-\frac 12)$ then we can check that $\|x\|=1$ and $f(x)=2$. Thus $\|f\| =2$.