Let $X=(C[0, \pi], \vert \vert . \vert \vert)$. for $f \in C[0, \pi]$ define $T(f)(x)=\int_0 ^x \sin (x+y)f(y)dy$. Find $\vert \vert T \vert \vert$
My answer is $3$. Is it correct?
$\vert \vert Tf \vert \vert_{\infty}=\sup\{\vert \int_0^x \sin(x+y)f(y)dy \vert\}=\sup\{\vert \int_0^x\sin x \cos y f(y)dy+\int_0^x\cos x \sin y f(y)dy\vert\}=\leq \sup\{\vert \int_0^x \cos y f(y)dy\vert\}+\sup\{\vert \int_0^x \sin y f(y)dy\vert\}\leq 3\vert \vert f \vert \vert_{\infty}$
Let $f \in X$ with $||f||_{\infty} = 1$ and $x \in [0, \pi]$. Then \begin{align*} |Tf(x)| &\leq \int_{0}^x |\operatorname{sin}(x+y)||f(y)|\text{ }dy \\ &\leq \int_{0}^{\pi} |\operatorname{sin}(x+y)|\text{ }dy \\ &=\int_{0}^{\pi} |\operatorname{sin}(y)|\text{ }dy \\ &= 2, \end{align*} where the second last equation is true because $|\operatorname{sin}|$ is $\pi$-periodic. So we have already shown $||Tf||_{\infty} \leq 2$ as well as $||T|| \leq 2$. To show the other inequality we consider the constant function $f \equiv 1$. Then obviously $||f||_{\infty} = 1$ and \begin{align*} |Tf(\pi)| &= \Bigg| \int_{0}^\pi \operatorname{sin}(\pi + y)\text{ }dy\Bigg| \\ &= |-2| \\ &= 2 \end{align*} and therefore $||Tf|| \geq 2$ which also prooves $||T|| \geq 2$.
So the result is $||T|| = 2$.