TITLE: Prove that $T$ is bounded and find its norm
QUESTION [5 upvotes]: Consider $T\colon C[0,1] \to C[0,1]$ by
$$T(f) = \int_{0}^1 \sin(t) \;f(t) \;dt$$

Show that $T$ is a bounded linear operator.
Find the norm of $T$.


This is my solution 

Is it true?
Thanks a lot..

REPLY [2 votes]: Vastly updated answer, providing a more elementary approach than the rather completely correct and elegant approach of Kavi Rama Murthy :
The linearity is easy to show (which you haven't done). Take $f, g \in C[0,1]$ and $\lambda \in \mathbb R$ and prove that $T(\lambda f+g) = \lambda Tf + Tg. $
For the boundedness, note that it is :
$$|Tx| = \bigg|\int_0^1 \sin(t)f(t) \mathrm{d}t \bigg| \leq \int_0^1 |\sin(t)f(t)|\mathrm{d}t  \leq \|f\|_\infty\int_0^1 |\sin(t)|\mathrm{d}t  $$
But, note that $\sin(t) \geq 0$ for $t \in [0,1]$, thus it is :
$$\|Tx\| \leq \|f\|_\infty\int_0^1\sin(t)\mathrm{d}t$$
That means that $T$ is a bounded linear operator $T : C[0,1] \to \mathbb R$ with $\|T\| \leq \int_0^1 \sin(t)\mathrm{d}t$.
Now, take $\mathbf{1} \in C[0,1]$. Then, it is :
$$T(\mathbf{1}) = \int_0^1\sin(t)\mathrm{d}t \implies \|T(\mathbf{1})\| = \int_0^1\sin(t)\mathrm{d}t \implies \|T\| = \int_0^1 \sin(t)\mathrm{d}t$$