Trace-Operator is compact?

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Assume $\Omega$ is a smooth, compact riemannian manifold with non-empty smooth boundary $\partial \Omega$. Let $T$ be the Trace-Operator $T \colon H^1(\Omega) \to L^2(\partial \Omega), \ f \mapsto f|_{\partial \Omega}$.

Is $T$ a compact operator? Using the edit, I'm asking now: Is $T$ a bounded operator?

Edit:

Since $\Omega$ is compact, we know that $\partial \Omega$ is also a compact manifold. So if we know that $T$ is bounded, we have already that $T$ is compact.