Compact integral and multiplication operator in Banach spaces Let $ A\colon C[0,1] \to C[0,1] $
$$ A(x)(t) = f(t)x(t) + \int_0^t x(s)ds,\quad f \in C[0,1]: f(1) \neq 0,  \forall t \in [0,1] $$
Is $A$ a compact operator or not?
 A: We should denote the operator $A_f$ in order to see the dependence on $f$. First, with Arzelà-Ascoli's theorem, show that the operator $J(x)(t):=\int_0^tx(s)ds$ is compact. So $A_f$ is compact if and only if so is the multiplication operator $x\mapsto xf$. This thread will give you ideas.
A: 
$A_\phi \colon C[0,1] \rightarrow C[0,1]$
$A_\phi (f)(x) = \phi (x) f(x), \forall f \in C[0,1], \forall x\in [0,1]$
$A_\phi $ is compact operator if and only if $ \phi \equiv 0 $.

If $ \phi \equiv 0 $ then $ A_\phi =0 $ is compact.
Conversely,let $ A_\phi $ is compact. $ B = \lbrace f \in C[0,1]: \Vert f \Vert \leqslant 1 \rbrace $. Then, $ A_\phi (B) $ is relatively compact in $ C[0,1] $. $ \forall a \in (0,1) $, choose $ n $ such that $ \dfrac{1}{n} < a $
Let $ f_n(x) = \left\lbrace 
\begin{array}{ll}
 0 & \mbox{if } 0 \leqslant x < a - \dfrac{1}{n};\\
 n(x-a) + 1 &\mbox{ if }a - \dfrac{1}{n} \leqslant x < a;\\
 1 & \mbox{if }a \leqslant x \leqslant 1.
\end{array} \right.$
Then, $ f_n $ is continuous and $ \Vert f \Vert \leqslant 1 $ so $ \phi\cdot f_n \in A_\phi (B) $. Since  $ A_\phi (B) $ is relatively compact in $ C[0,1] $, by Arzela - Ascoli's theorem , $ A_\phi (B) $ is local equicontinuous, so $ \forall \varepsilon > 0, \exists \delta > 0, \forall x, y \in [0,1], \vert x -y \vert < \delta $, we have $ \vert f_n(x) \phi (x) - f_n(y) \phi(y) \vert < \varepsilon, \forall n \geqslant \dfrac{1}{a}. $
Choose $ n $ such that $ n >\max \lbrace \dfrac{1}{a}, \dfrac{1}{\delta} $. Then, if $ x = a - \dfrac{1}{n}, y = a $, $  \vert x - y \vert = \dfrac{1}{n} < \delta $ and $ f_n(x) =0, f_n(y) =1 $. Therefore, $ \phi (a) =0, \forall a \in (0,1) $. By continuity of $ \phi $, $ \phi (x) = 0, \forall x \in [0,1] $.
A: How about taking $f$ to be the constant function with value $1$? Then $A=I+V$ where $V$ is a Volterra operator (compact), and $A$ doesn't look compact to me.
