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I was checking nice theorems on wikipedia and I stumble on this one (BLT theorem): https://en.wikipedia.org/wiki/Continuous_linear_extension

It says that every bounded linear transformation $T$ from a normed vector space $V$ to a complete, normed vector space $W$ can be uniquely extended to a bounded linear transformation ${\tilde T}$ from $\bar V$, the closure of $V$, to $W$.

The article cites Reed's book, but there is no BLT theorem in this one. I usually check the proofs after the wikipedia page, but I cannot find any trace of this theorem anywhere. Does someone know a reference for this ?

Thank you

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An Introduction to Banach Space Theory, Robert E. Megginson, page 70.

The crucial step is to justify the well-definedness of the bounded linear operator $\overline{T}:\overline{V}\rightarrow W$ defined by $\overline{T}(v)=\lim_{n\rightarrow\infty}T(v_{n})$ where $(v_{n})\subseteq V$ is such that $v_{n}\rightarrow v$.

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This result is true in a more general context:

Theorem. Let $X$ and $Y$ be metric spaces, $A$ a dense subset of $X$ and $f\colon A\rightarrow Y$ a uniformly continuous map. If $Y$ is complete, then there exists a unique continuous map that extends $f$ to $X$.

Sketch of a proof. Let $a\in A$, there exists $(a_n)_n$ a sequence of elements of $A$ converging to $a$, then $(f(a_n))_n$ is a Cauchy sequence in $Y$, which is complete, so that it converges to a point $y\in Y$. The key point is that $y$ does not depend on the sequence $(a_n)_n$ but just on $a$, so that $\widetilde{f}(a):=y$ is a valid definition. The rest of the proof consists in seeing that $\widetilde{f}$ is (uniformly) continuous, which should be clear.

The uniqueness part follows from the fact that two continuous map agreeing on a dense subset are equal. $\Box$

A continuous linear map is Lipschitz so that it is uniformly continuous and the above result can be applied, notice that the process used to defined $\widetilde{f}$ is linear.

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  • $\begingroup$ Completeness and uniform continuity aren't topological notions. You need uniformities to define them. Moreover, in this generality you cannot argue with sequences. $\endgroup$ – Jochen Feb 13 '18 at 9:04
  • $\begingroup$ @Jochen I assumed it was clear that when I say that $f$ is uniformly continuous, then both $X$ and $Y$ are endowed with metrics because as you said, this makes no sense otherwise. Anyways, I edited to make it explicit! $\endgroup$ – C. Falcon Feb 13 '18 at 14:03

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