Examples of compact operators, solution verification This is an exercise in the "lecture notes in functional analysis" by bressan. asking to show whether the following operators are bounded and compact on the Banach space $X = C([0, 1])$, 
(I try to write all the detail steps)
$\textbf{- $\Lambda f(x)=f(sinx)$ }$


*

*bdd : $|\Lambda f(x)|=|f(sinx)| \le \|f\| \implies \|\Lambda f\| =  \|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} =  \frac{{\|f\|}}{{\|f\|}} =1 $

*not compact : let $f_n(x) =x^n$ , then $\Lambda f_n(x)= [sin(x)]^n$ , if $sin(x)=1 , \Lambda f_n(x) \to 1$ , otherwise $\ \ \Lambda f_n(x) \to 0 $. So there would be no converging subsequence for this example. 
$\textbf{-$\Lambda f(x)=xf(x)$}$


*

*bdd : $|\Lambda f(x)|=|xf(x)| \le \|x\|\|f\| \implies \|\Lambda f\| = \sup_{x}\frac{\|x\|\|f\| }{\|x\|}=\|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} =  \frac{{\|f\|}}{{\|f\|}} =1 $

*not compact : let $f_n(x) =x^n$ , then $\Lambda f_n(x)= x^{n+1}$ , if $x=1 , \Lambda f_n(x) \to 1$ , otherwise $\ \ \Lambda f_n(x) \to 0 $. So there would be no converging subsequence for this example. 
$\textbf{-$\Lambda f(x)=xf(0)+\int_0^1 f(s)ds$}$


*

*bdd : $|\Lambda f(x)|=|xf(0)+\int_0^1 f(s)ds|\le 2\|f\| \implies \|\Lambda f\| =  2\|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} =  \frac{2\|f\|}{\|f\|} =2 $

*compact : we always have $\Lambda f(x)=xf(0)+\int_0^1 f(s)ds = P(\text{at most degree 1})+\text{constant}$ , so the range of $\Lambda f(x)$ is at most two. so it is a compact operator.
$\textbf{-$\Lambda f(x)= y(x)$}$ ,  where y(·) is the solution to the Cauchy problem $y'(x)+y(x)=f(x) , \ \ y(0)=0$


*

*bdd : we have that by above $\Lambda f(x) = \int_0^x e^{y-x}f(y)dy \implies |\Lambda f(x)| = |\int_0^x e^{y-x}f(y)dy| \le \|f\| \implies \|\Lambda\| = 1 $

*compact :by the compactness of an integral operator.
 A: For the first one, $\Lambda f(x)=f(\sin x)$. Note that when $x\in[0,1]$, we have $\sin x\in[0,\sin1]\subset[0,1]$. This is because $1<\pi/2$ so $\sin 1<1$. As a result, for any $f$, it follows that $\sup_{x\in[0,\sin 1]}|f(x)|\leq \sup_{x\in[0,1]}|f(x)|=||f||.$
Bounded: I agree that $|\Lambda f(x)|\leq ||f||$ for all $x\in [0,1]$ so $||\Lambda f||\leq  ||f|| \implies ||\Lambda||\leq 1$. This, of course, finishes the proof. But in your answer, you claim that $||\Lambda||=1$. You really should show that there exists $f$ such that $||\Lambda f||=||f||$. This is easy, as just choose $f=1_{[0,1]}$.
(not) Compact: Your counterexample doesn't prove it. This is because for $f_n(x)=x^n$, we have $\Lambda f_n(x)=(\sin x)^n$. Therefore, $||\Lambda f_n||= (\sin 1)^n \to 0$ since $\sin 1<1$. Hence, obviously, a convergent subsequence exists. There is an easy way to modify your counterexample to make your main idea hold true.
For the second one, $\Lambda f(x)=xf(x)$, you can say $|\Lambda f(x)|\leq |x|||f|| \leq ||f||$ so $||\Lambda|| \leq 1$. For the other inequality, choose $f(x)=1$ again. This time I agree with your proof of (not) compactness.
In general, you keep making similar mistakes. I suggest you read more about how to prove the norm of an operator equals a number.
