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Let $X$ be a Banach space. Show that any nontrivial $f \in B(X,\mathbb{K})$ is an open map.

This is a simple question, but I don't see how it is solved. If $f$ is surjective, then we can apply the Open Mapping Thm to show that $f$ is open. But how to see that $f$ is surjective?

José Carlos Santos
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TheMauritsD
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1 Answers1

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I suppose that “non-trivial” means that it's not the null map. But then $f(X)$ contains some non-null scalar from $\mathbb K$ and, since it must be a vector subspace of $\mathbb K$, it must be the whole $\mathbb K$.

José Carlos Santos
  • 427,504