Find $\Vert f \Vert$ for a functional $f$ over $C[-1,1]$.

Question

Let , $\displaystyle f:C[-1,1]\to C[-1,1]$ be defined by $\displaystyle f(x)=\int_{-1}^0 x(t)\,dt -\int_0^1x(t)\,dt$. Find $\Vert f\Vert$ with respect to sup norm defined on $C[-1,1]$.

Firstly , $\displaystyle |f(x)|\le \int_{-1}^0\sup_{-1\le t\le 1}|x(t)|\,dt+\int_0^1 \sup_{_1\le t\le 1}|x(t)|\,dt=2\Vert x\Vert$.

Then we get , $\Vert f\Vert \le 2$.

I couldn't find the reverse inequalty. I'm unable to construct $\Vert f\Vert \ge 2$.

I saw this but that answer is not clear to me.

Answer 1

Consider the sequence of functions

$$x_n(t) = \left\{ \begin{array} .1 & \forall t \in [-1,-\frac{1}{n}] \\ - nt & \forall t \in [-\frac{1}{n},\frac{1}{n}] \\ -1 & \forall t \in [\frac{1}{n},1] \end{array} \right.$$

The $x_n$ are continuous, $\|x_n\| = 1$, and $f(x_n) \to 2$ when $n\to\infty$

Answer 2

Another example: Let $x_n(t) = t^{1/(2n+1)}, n = 1,2,\dots $ Then $\|f_n\|= 1$ for every $n,$ while

$$f(x_n) = \frac{2}{1+ 1/(2n+1)} \to 2.$$