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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?2012-09-09
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    I mean that they are equivalent. The space is the space of smooth functions2012-09-09
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    I don't understand: in the title you write equivalence, and in the body of the question you ask the two norms to agree. Do you mean the two norm agree (or are equivalent) for the functions whose support is contained in a compactly contained open set?2012-09-09
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    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.2012-09-09
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    In fact, I'm now wondering if infinite-dimensional Banach spaces actually *have* any compactly contained open sets...2012-09-09
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    @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.2012-09-09
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    I edited the original post, I hope it is now clear2012-09-09
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    No, it is not clear what you mean by “аre equivalent in an open set”.2012-09-09
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    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?2012-09-09
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    In which sense is $\lbrace f: V\to \mathbb R: f \text{smooth}\rbrace$ a subspace of $\lbrace f: U\to \mathbb R: f \text{smooth}\rbrace$? Do you mean instead $\lbrace f: U\to \mathbb R: \text{support}(f) \subseteq V \rbrace$?2012-09-10
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    @Jochen I don't think it is important since with cut off functions you can extend the functions defined on $V$ to the whole domain $U$.2012-09-11

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