Let $X$ and $Y$ be normed spaces, both real or both complex; let, in addition, $Y$ be a Banach space; let $V$ be a (vector) subspace of $X$; let $T \colon V \to Y$ be a bounded linear operator; and let $\overline{V}$ denote the closure in $X$ of $V$. Then there exists a unique bounded linear operator $S \colon \overline{V} \to Y$ such that $$S(v) = T(v) \ \mbox{ for all } \ v \in V$$ and $$\Vert S \Vert = \Vert T \Vert.$$

This is Theorem 2.7-11 in Introductory Functional Analysis With Applications by Erwine Kreyszig.

What is the name, if any, for this theorem in the standard literature on functional analysis?

  • $\begingroup$ Quoting this wikipedia article - "This theorem is sometimes called the BLT theorem, where BLT stands for bounded linear transformation." $\endgroup$ – stochasticboy321 Dec 4 '15 at 5:28

As stated, the theorem is wrong, take e.g a Banach space $X$, a dense subspace $V$ and consider the identity $V\to V$. For complete $Y$ it is indeed true but has little to do with linearity. One could call it extension principle for uniformly continuous maps which holds for a metric space $X$ and a complete metric space $Y$ (or, if you like, uniform and complete uniform spaces): Every uniformly continuous map from a subset of $X$ to $Y$ has a unique continuous extension to the closure (which is again uniformly continuous).

  • $\begingroup$ you're absolutely right about the case when $Y$ is not complete. However, the whole point of linearity is that then the extension is also linear. Moreover, for linear operators, boundedness is synonymous with uniform continuity. $\endgroup$ – Saaqib Mahmood Dec 5 '15 at 11:29

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