Let $T:L^{1}(-\infty,\infty)\rightarrow C(-\infty,\infty)$ be an operator defined by $$(Tf)(x)=\int_{-\infty}^{x}f(t)\sin(t)dt$$ Show that is is bounded and calculate its norm. The space $C(-\infty,\infty)$ has a norm $\left\Vert f\right\Vert _{\infty}=\sup_{t\in(-\infty,\infty)}\left|f(t)\right|$ while $L^{1}(-\infty,\infty)$ has a norm $\left\Vert f\right\Vert=\int_{-\infty}^{\infty}|f(t)|dt$.
An attempt to solve
I try to check the condition $\Vert Tf \Vert \leq C\Vert f\Vert$ fulfilled for all $f \in L^1(-\infty,\infty)$. Now I get confused because I think I should apply the norm in $L^1$ space for this inequality, but I just feel I don't know how and that I'm doing it completely wrong.
For the norm of the operator, I apply $\left\Vert T\right\Vert =\sup_{\left\Vert f\right\Vert=1}\Vert Tf \Vert$ and again it would seem the norm is taken in $L^1$ space.
If I would go for the norm defined in $C$ space I could analyze supremum, but essentially I investigate the value of the integral which is bounded $$\int_{-\infty}^{x}f(t)\sin(t)dt\leq\sup_{t\in(-\infty,\infty)}\left|f(t)\right|\cdot\int_{-\infty}^{x}\sin(t)dx$$ and since the sine integral is less or equal 1 if we could even calculate it because of the infinity, we have a bound. I feel like I mixed up some concept severely here.
update 1 So we could go, like the comment below suggest, somewhere there: $$\left\Vert Tf\right\Vert _{\infty}=\sup_{x\in\mathbb{R}}\left|\int_{-\infty}^{x}f(t)\sin(t)dt\right|\leq\sup_{x\in\mathbb{R}}\int_{-\infty}^{x}\left|f(t)\sin(t)\right|dt\leq \\ \leq \sup_{x\in\mathbb{R}}\int_{-\infty}^{x}\left|f(t)\right|dt \leq\int_{-\infty}^{\infty}\left|f(t)\right|dt= \left\Vert f\right\Vert _{1}$$ which would lead to $\left\Vert Tf\right\Vert _{\infty}\leq\left\Vert f\right\Vert _{1}$. Even if I would go that way, what about the norm? How to even use $\left\Vert f\right\Vert _{1}=1$?