# 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?

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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.

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ok. $x \rightarrow xf$ is compact operator if and only if $f \equiv 0$ –  daisy Dec 13 '12 at 11:04

$A_\phi \colon C[0,1] \rightarrow C[0,1]$

$A_\phi (f)(x) = \phi (x) f(x), \forall f \in C[0,1], \forall x\in [0,1]$

$A_\phi$ 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 $\phi$, $\phi (x) = 0, \forall x \in [0,1]$.

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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.