Is the integral operator with continuous kernel from $H^{-\frac 1 2}(\partial\Omega)$ to $H^{\frac 1 2}(\partial\Omega)$ a compact operator?

Question

For a bounded domain $\Omega$ with $C^1$ boundary, define a integral operator $A$ with a continuous kernel $k$ satisfies $$ A f (x) = \int_{\partial\Omega} k(x,y)f(y)~{\rm d}y \quad {\rm for }~x\in\partial\Omega. $$ Here $k(x,y)$ is continuous on $\partial\Omega \times \partial\Omega$.

I know $A:L^2(\partial\Omega) \rightarrow L^2(\partial\Omega)$ is compact. Then is $A:H^{-\frac 1 2}(\partial\Omega) \rightarrow H^{\frac 1 2}(\partial\Omega)$ still compact?