$C[-1,1] $ above $$ f(x)=\int_{-1}^{0}x(t)dt-\int_{0}^{1}x(t)dt$$ What is norm of the f linear fucntional?
I tried to solve using definition of norm but I couldn't find result. Please can you give the solution or hints?
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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 $\Vert f \Vert$ for a functional $f$ over $C[-1,1]$, Finding the bound of a linear functional defined on $C[-1,1]$. – Martin Sleziak Jan 05 '18 at 03:45
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Consider $C[-1,1]$ with the supremum norm $\|\cdot\|_\infty$. Then $|f(x)|\le \int\limits_{-1}^0|x(t)|dt+\int\limits_{0}^1|x(t)|dt\le 2\|x\|_\infty,\,\forall x\in C[-1,1].$ Thus $\|f\|=\sup\limits_{{x\neq 0},{x\in C[-1,1]}}\frac{|f(x)|}{\|x\|_\infty}\le 2$. To show that $\|f\|$ is actually $2$, for each positive integer $n$, consider the continuous function $x_n:C[-1,1]\to \Bbb R$ given by $$x_n(t)= \begin{cases} 1\quad &-1\le t\le-\frac{1}{n}\\ -nt\quad &-\frac{1}{n}\le t\le \frac{1}{n}\\ -1\quad &\frac{1}{n}\le t\le 1. \end{cases}$$ Then $$f(x_n)=\int\limits_{-1}^{-\frac{1}{n}}dt+\int\limits_{-\frac{1}{n}}^{0}{(-nt)dt}-\left(\int\limits_{0}^{\frac{1}{n}}{(-nt)dt}+\int\limits_{\frac{1}{n}}^{1}(-1)dt\right)=2-\frac{1}{n}.$$ Also $\|x_n\|_\infty=1,\quad\forall n\in\Bbb N.$ Hence using the definition of $\|f\|$, we have $\|f\|\ge \frac{|f(x_n)|}{\|x_n\|_\infty}=2-\frac{1}{n},\, \forall n\in \Bbb N.$ So we have, $\|f\|\ge 2$. Hence combining these two inequalities we have $\|f\|=2$.
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