Let $\Omega \subset \ R$ be bounded and regular domain and \begin{array}{ccccc} \nabla :& H^1(\Omega) & \longrightarrow &L^2(\Omega)\\ & y & \longmapsto & \displaystyle \frac{\partial y(x)}{\partial x} \end{array} Is the following operator compact
\begin{array}{ccccc} K :& H^1(\Omega) & \longrightarrow & L^2(\Omega)\\ & y & \longmapsto & \displaystyle\int_0^x\nabla y(s)ds \end{array}
For $n=1$ a function in $H^1(\Omega)$ is absolutely continuous, so you can apply the fundamental theorem of calculus to get that $$\int_0^xy'(s)\,ds=y(x)-y(0).$$ If you have a bounded sequence $\{y_n\}_n$ in $H^1(\Omega)$ then by the Ascoli-Arzela theorem a subsequence $\{y_{n_j}\}_{j}$ will converge uniformly. So yes, the operator $K$ is compact since by uniform convergence $y_{n_j}(x)-y(0)\to y(x)-y(0)$.