# Linear transformation bounded iff its kernel is closed in infinite-dimensional Banach spaces

I am working on Problem 8, Chapter 6, in Luenberger's Optimization by Vector Space Methods. It states:

"Show that a linear transformation mapping one Banach space into another is bounded if and only if its nullspace is closed."

I am having a bit of trouble with the converse. In particular, if we let $$f:X \rightarrow Y$$ be a linear transformation, Luenberger doesn't assume that either $$X$$ or $$Y$$ be finite-dimensional. Do you have any idea on how to proceed? I have thought (without success) of considering the quotient space $$\hat{X}/\ker f$$

Let $$X$$ be any infinite-dimensional banach space with (Hamel) basis $$B$$ and let $$\{b_n\}_{n\in\ \mathbb N}$$ be a countable subset of $$B$$. Then define $$T:X\to X$$ on the basis $$B$$ by $$T(b_n)=nb_n$$ and $$T(b)=b$$ for $$b\in B\setminus \{b_n\}_{n\in\ \mathbb N}$$ and extend linearly. Then $$T$$ is an unbounded operator, with $$\ker T=\{0\}$$ closed.
• @PeterFranek: Sure it is well defined. Since $B$ is a Hamel basis, every vector in $X$ can be uniquely written as a finite linear combination of the elements of $B$. Commented Jun 6, 2020 at 23:11
• Thanks a lot! I will accept this answer. I have one additional question however: is it possible to construct the $B$ in this counterexample without using the axiom of choice? Commented Jun 6, 2020 at 23:14