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.

Try to construct a function $x$ with$ |x| \le 1$ to make $f(x) = 2$. Or (since $x$ has to be continuous) nearly $2$. — GEdgar, Aug 01 '16 at 14:46
Your answer depends on you define the norm $||\cdot||$ on the space ${\cal C}[-1, 1]$. — avs, Aug 01 '16 at 14:47
See also: Norm of the operator $Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt$, Finding the bound of a linear functional defined on $C[-1,1]$, find linear functional norm, Finding the bound of a linear functional defined on $C[-1,1]$. — Martin Sleziak, Jan 05 '18 at 03:44

score 2 · Accepted Answer · answered Aug 01 '16 at 15:03

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 $n\to\infty$

zhw. · Answer 2 · 2016-08-01T18:05:03.907

1

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

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

edited Aug 01 '16 at 18:05

answered Aug 01 '16 at 15:20

zhw.

105,693

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

2 Answers2

Linked