General practice to find the operator norm I usually have problem to find the operator norm. Take the following example for instance;
Let $H=L^2([0,\pi])$ and $\Lambda:H\to H$ such that $\Lambda f(x) = f(x) \sin(x)$, I want to find the operator norm. 
Usually it is that we have to find two inequalities. I have mostly found that using the norm definition we can have one of the inequalities. In particular, I can proceed as the following for one of them. 
$$\|\Lambda f\|_{L^2([0,\pi])}^2= \int_0^\pi |f(x)\sin(x)|^2 \le \int_0^\pi |f(x)|^2 = \|f\|_{L^2([0,\pi])}^2  \implies \|\Lambda\| \le 1$$
But I mostly have no clue to proceed how to get the other inequality. So I wondered if I can have  more general practice(s) that I have to follow or check (depending on the question).
 A: For the "other inequality", ideally you are looking for a function such that $\|\Lambda f\|=\|f\|$. But that's often not the case. So the next best thing is to get as close to 1 as you can. Here, the sine function is 1 at $\pi/2$, and -1 at $3\pi/2$. Anyway else, its value is less than 1 in absolute value so it will be "making $f$ smaller". 
So what you want to do is consider $f$ concentrated on $\pi/2$. If you consider $f=1_{[\tfrac\pi2-\tfrac1n,\tfrac\pi2+\tfrac1n]}$, then 
$$
\|\Lambda f\|^2=\int_{\pi/2-1/n}^{\pi/2+1/n}|\sin(t)|^2\,dt=\tfrac1n+\sin\tfrac1n\cos\tfrac1n.
$$
Then norm of $f$ is $\sqrt{\tfrac2n}$. So 
$$
\frac{\|\Lambda f\|^2}{\|f\|^2}=\frac{\tfrac1n+\sin\tfrac1n\cos\tfrac1n}{\tfrac2n}
=\tfrac12\left(1+\frac{\sin\tfrac1n\cos\tfrac1n}{\tfrac1n}\right).
$$
The fraction inside the brackets can be made as close to 1 as desired. So $\|\Lambda|\geq1-\varepsilon$ for all $\varepsilon>0$, which implies that $\|\Lambda\|\geq1$. 
I don't think there is a general answer to your question. Different operators are expressed in very different ways, so the ideas to find the norm can be very very different. 
