Question

Find the norm of the linear functional f defined by

𝑓 𝑥 =integral -1 to 0𝑥 𝑡 𝑑𝑡 − integral 0 to 1𝑥 𝑡 𝑑𝑡

where𝑥 ∈ [−1 , 1]

Answer 1

$|f(x)|=$$|\int^0_{-1}x(t)dt-\int^1_{0}x(t)dt|\le |\int^0_{-1}x(t)dt|+|\int^1_{0}x(t)dt|$

$|\int^0_{-1}x(t)dt|+|\int^1_{0}x(t)dt|\le {\sup}_{t\isin [-1,0]}|x(t)|+{\sup}_{t\isin [0,1]}|x(t)|\le 2{\sup}_{t\isin [-1,1]}|x(t)|=$

$=2||x(t)||_{\infin}$

thus, $||f||\le2$

if we consider the sequence

$x_n^*(t)=\begin{cases} 1 &\text{if } -1\le t\le -1/n \\ -nt &\text{if } -1/n< t< 1/n \\ -1 &\text{if } 1/n\le t\le 1 \end{cases}$

$x_n^*(t)\isin C[-1,1]$ for each $n\isin N$

Then we see that

$||f||=sup_{x\in C[-1,1]:||x||=1}|f(x)|\ge |f(x_n^*)|\forall n \isin N$

since $|f(x_n^*)|=|2+1/n^2-1/n|$ :

$||f||\ge |2+1/n^2-1/n| \implies ||f|| \ge 2$

so, $||f|| =2$