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"Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on the space $X$ s.t. $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$ are both complete. Assume that for any sequence $(x_n) \subseteq X$, $\|x_n\|_1 \to 0$ always implies $\|x_n\|_2 \to 0$. Prove that the two norms are equivalent, i.e. $\exists a,b>0$ s.t. $\forall x \in X$, $a\|x\|_1 \leq \|x\|_2 \leq b\|x\|_1$."

This is a problem from Erwin Kreyszig's "Introductory Functional Analysis with Applications". Sorry for the vagueness of the title; I couldn't think of an adequate title.

This problem was given as an exercise right after the section on the Open Mapping Theorem so I'm thinking that's what the author would like the reader to use. However, I don't see any connection.

If you have an idea on how OMT is related to the problem or a solution, that would be great.

Thanks in advance.

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Consider the identity mapping $I: (X, \|\cdot\|_1) \rightarrow (X, \|\cdot\|_2)$, your given condition implies $I$ is continuous, so $\|x\|_2 \leq C\|x\|_1$

Since $I(X) = X$ is of second category in $X$, by open mapping theorem $I$ is an open mapping, which means $I^{-1} = I$ is continuous from $(X, \|\cdot\|_2)$ to $(X, \|\cdot\|_1)$, so $\|x\|_1 \leq C'\|x\|_2$

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Consider the (linear) identity map $$I:(X,||\cdot||_{1})\to(X,||\cdot||_{2}).$$

Your condition implies $I$ is continuous. By the open mapping theorem, $I$ is an open map, and thus a homeomorphism since the identity is bijective. This implies the inverse is continuous, hence the equivalence of norms.

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The identity $i : X_1 \to X_2$ where $X_1 = (X, || \cdot||_1)$ and $X_2 = (X, || \cdot||_2)$ is a bounded linear operator on Banach spaces, thus is an oppen mapping, and because of the Open Mapping theorem it has a continuous (bounded) inverse. This implies that the norms are equivalent. Why, because:

$$||x||_2 = ||i(x)||_2 \leq C||x||_1$$

And $$||i^{-1}x||_1 = ||i(x)||_1 \leq M||x||_2$$

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