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

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 $$\ker f$$?

• $\operatorname{ker}(f)=\{x\in C([0,1],\mathbb{R}):\ x(0)=x(1)\}$ is a way to write it. Commented Jan 17, 2020 at 13:25

$$\ker (f)= \{x\in C[0,1]: x(0)=x(1)\}$$.
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 $$\|f\| =2$$.