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Ayman Moussa
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If the setting is the one of my comment above, I think that $C^k_0(\Omega)$ is not Montel, for finite $k$. Below, for $K\subset \Omega$ a compact set I write $C^k_K(\Omega)$ the set of all $C^k(\Omega)$ functions having support included in $K$. I also fix a countable exhaustion $(K_n)_n$ of $\Omega$.

One property (characterization) of the inductive limit topology on $C^k_0(\Omega)$ is the following : it's the finest LCTVS topology for which the embeddings $C^k_{K_n}(\Omega)\hookrightarrow C^k_0(\Omega)$ are all continuous. But $C^k_{K_n}(\Omega)$ is a normed space of infinite dimension : it cannot be Montel (this arguments fails when $k=\infty$ obviously), for instance the unit ball of $C^k_{K_n}(\Omega)$ is bounded and not compact. Thanks to the previous embedding, this unit ball is also bounded in $C^k_0(\Omega)$ and cannot be compact in it, otherwise this would imply compactness in $C^k_{K_n}(\Omega)$.

Ayman Moussa
  • 3.4k
  • 1
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