# equivalence of norms in an open set

I would like to prove that two given norms in the space of smooth functions are equivalent in an open set, is it enough to show that they are equivalent for any compactly contained open set? why?

Edit: Clarification promoted from the comments: Consider the space $$\{f\colon U \to \mathbb{R}:f \text{ is smooth}\}$$ and consider two given norms on this space. Is it enough in order to prove that these two norms are equivalent that in any of the spaces $$\{f\colon V \to \mathbb{R} : f \text{ is smooth}\}$$ where $V$ is compactly contained in $U$, the two norms are equivalent?

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What do you mean by "two norms agree in an open set"? In which space are you working? – Davide Giraudo Sep 9 '12 at 10:16
I mean that they are equivalent. The space is the space of smooth functions – inquisitor Sep 9 '12 at 10:25
I'd be a little surprised if it was enough. Note that if a space is not locally compact then some points are not a member of any compactly contained open set. – Ben Millwood Sep 9 '12 at 10:38
@BenMillwood: The answer to your question is no, because of Riesz's lemma. But this is OT unless the OP clarifies her(his) question, which makes no sense at the moment. – Giuseppe Negro Sep 9 '12 at 10:45
Consider the space $$\{f: U \rightarrow R : f smooth\}$$ and consider two given norms on this space. Is it enough in order to prove that these two norms are equivalent that in any of the spaces $$\{f: V \rightarrow R : f smooth\}$$ where $V$ is compactly contained in $U$, the two norms are equivalent? – inquisitor Sep 9 '12 at 18:03