Let $C[0,1]$ denote the set of all continious real valued function on the interval $[0,1]$ and $T:(C[0,1],||.||_{\infty})\to \mathbb R$ be defined by $T(f)=\int_0^1tf(t)$ for all $f \in C[0,1]$. Then find $||T||$?
My attempt:
i take constant function $f(t)=1$ i got $||T||=1$.
is its true ???
Any hints/solution
thanks u
For any $f\in C[0, 1]$, we have $$ |T(f)| = \Big|\int_{0}^{1}tf(t)dt\Big| \leq \int_{0}^{1}t |f(t)|dt \leq ||f||_{\infty}\int_{0}^{1}tdt = \frac{1}{2}||f||_{\infty} $$ so $||T||\leq 1/2$, and we have $||T||\geq 1/2$ by taking $f \equiv 1$.
Well, note that
$\vert T(f) \vert = \left \vert \displaystyle \int_0^1 tf(t) \; dt \right \vert \le \displaystyle \int_0^1 t \vert f(t) \vert \; dt \le \displaystyle \int_0^1 t \Vert f \Vert_\infty \; dt = \Vert f \Vert_\infty \int_0^1 t \; dt = \dfrac{1}{2} \Vert f \Vert_\infty, \tag 1$
which shows that
$\Vert T \Vert \le \dfrac{1}{2}; \tag 2$
now take
$f(t) = 1, \tag 3$
and find that
$\vert T(1) \vert = \left \vert \displaystyle \int_0^1 t \; dt \right \vert = \dfrac{1}{2} = \dfrac{1}{2} \Vert 1 \Vert_\infty, \tag 4$
which precludes the possibility that
$\Vert T \Vert < \dfrac{1}{2} \tag 5$
since this would imply
$\vert T(1) \vert \le \Vert T \Vert \Vert 1 \Vert_\infty < \dfrac{1}{2} \Vert 1 \Vert_\infty, \tag 6$
contradicting (4). Therefore,
thus
$\Vert T \Vert = \dfrac{1}{2}. \tag 7$
