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).

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.