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Suppose that $X$ is an infinite-dimensional normed vector space over $\mathbb C$ (that is, the cardinality of any of its Hamel bases is infinite) and let $Y$ be another, nontrivial normed vector space over $\mathbb C$ (that is, $Y$ contains other elements beyond the zero vector). I am wondering about the following questions:

  • If $X$ is incomplete (that is, there exists at least one Cauchy sequence in $X$ that fails to converge in $X$), how can one construct a linear function from $X$ to $Y$ that is unbounded (or, equivalently, not continuous) explicitly, i.e., without using the axiom of choice or other results hinging on it (e.g., the Hahn–Banach theorem)?
  • If $X$ is complete, how can one construct an incomplete subspace $X_0$ of it such that the completion of $X_0$ is precisely $X$?

I appreciate any hints or references to standard results.

triple_sec
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1 Answers1

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Proving the existence of Schauder bases sometimes relies on the axiom of choice, but many important normed spaces have explicit Schauder bases.

Suppose we know a (countable) Schauder basis for $X$, call it $\{x_i\}$. The subspace $X_0$ for which $\{x_i\}$ is a Hamel basis is incomplete. Explicitly $X_0=\{\sum_{j=1}^Nc_jx_{i_j}:N\in\mathbb{N},c_j\in\mathbb{C}\}$.

To show $X_0$ is incomplete, notice that the sequence $\left\{\sum_{j=1}^Nx_j2^{-j}\right\}_{N=1}^\infty$ is Cauchy but does not converge in $X_0$. It shouldn't be hard to show $X$ is the completion of $X_0$ (basically that's how a Schauder basis is defined).

You can construct an unbounded linear functional $f:X_0\to\mathbb{C}$ easily: $f(\sum_{j=1}^Nc_jx_{i_j})=\sum_{j=1}^Ni_jc_j$.

To show $f$ is unbounded, assume $\|x_i\|=1$ for each $i$. Then $f(x_i)=i$ and so $\{x_i\}$ is a sequence on the unit sphere with an unbounded image.

If $0\neq y\in Y$, you can define an unbounded linear function $g:X_0\to Y$ by $g(x)=yf(x)$.

Babamots
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