Find $\ker(f)$ and norm of linear functional in Banach space. In Banach space $C([-\pi,\pi],\mathbb{R})$ we have linear functional $f:C([-\pi,\pi], \mathbb{R})\rightarrow \mathbb{R}$ such as $f(x)=\int_{-\pi}^{\pi} x(t)\operatorname{sign}(\sin(t))\,dt$ for any $x \in C([-\pi,\pi],\mathbb{R})$.
Calculate $\ker(f)$, $d(t,\ker(f))$, and its norm $\|f\|$.
Would appreciate for tips on how to get started. Tried to compute $\ker f$ but sign function scares me.
 A: By definition of $\text{sign}$-function we have $\text{sign}(\sin t)=1$ at $0<t<\pi$ and $\text{sign}(\sin t)=-1$ at $-\pi<t<0$, so functional can be rewritten as $f(x)=\displaystyle\int\limits_0^{\pi}x(t)dt-\displaystyle\int\limits_{-\pi}^0x(t)dt$. Thus we have $$\ker f=\left\{x\in C[-\pi,\pi]:\;\displaystyle\int\limits_0^{\pi}x(t)dt=\displaystyle\int\limits_{-\pi}^0x(t)dt\right\}.$$
The value of $d(t,\ker f)$ is calculated by the formula $d(t,\ker f)=\dfrac{|f(t)|}{\|f\|}$. Find the norm $f$: $|f(x)|\leq\displaystyle\int\limits_0^{\pi}|x(t)|dt+\displaystyle\int\limits_{-\pi}^0|x(t)|dt\leq\|x\|\cdot\left(\displaystyle\int\limits_0^{\pi}1dt+\displaystyle\int\limits_{-\pi}^01dt\right)=2\pi\|x\|$, so $\|f\|\leq2\pi$. If we consider the sequence of functions $x_n(t)=1$ at $\frac{1}{n}\leq t\leq\pi$, $x_n(t)=-1$ at $-\pi\leq t\leq-\frac{1}{n}$, and $x_n(t)$ is linear on the rest part $[-\frac{1}{n}, \frac{1}{n}]$, then we get that $\|f\|\geq\dfrac{|f(x_n)|}{\|x_n\|}=|f(x_n)|\to 2\pi$ at $n\to \infty$. Thus $\|f\|=2\pi$.
