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Suppose $ X $ is a Banach space with respect to two different norms, $ \|\cdot\|_1 \mathrm{ e } \|\cdot\|_2 $. Suppose there is a constant $ K > 0 $ such that $$ \forall x \in X, \|x\|_1 \leq K\|x\|_2 .$$ show then that these two norms are equivalent

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This is a variant of the open mapping theorem. If we consider the identity map $i$ on $X$ as a linear mapping from $(X, \|\cdot\|_2)$ to $(X, \|\cdot\|_1)$, then your condition says that $i$ is continuous. Then by the theorem $i$ is open and hence a homeomorphism, so its inverse is also continuous and the norms are equivalent.

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  • $\begingroup$ Beautiful!! Thanks! $\endgroup$
    – 6005
    May 14 '14 at 6:14
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Use the open mapping principle: the identity map $F: (X, \Vert \cdot \Vert_2) \to (X, \Vert \cdot \Vert_1)$ is linear and continuous (thanks to hypothesis). Its rank is of the second category, since $X$ is Banach (in particular, a complete metric space under the metric induced by $\Vert \cdot \Vert_1$).

By open mapping, it follows the this is an open map, i.e. an homeomorphism. In particular, the inverse map $F^{-1}: (X, \Vert \cdot \Vert_1) \to (X, \Vert \cdot \Vert_2)$ is still continuous; from this you easily get the other inequality you need in order to prove that the norms are equivalent.

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