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Let $X$ and $Y$ be locally convex topological vector spaces, say over $\mathbb{C}$. To set the stage a bit, I'll say that the topology on $X$ is given by a separating family of semi-norms $(p_i)_{i \in I}$ and, similarly, the topology on $Y$ is given by a family $(q_j)_{j \in J}$.

Now, suppose that $V \subset X$ is a dense subspace of $X$ and $T : V \to Y$ is a continuous linear map. Does there exist, then, a (unique) continuous extension $\overline T : X \to Y$ of $T$?

The answer is yes for Banach spaces, and this is an incredibly useful fact. The norm seems pretty crucial to the proof though. Any easy counterexamples?

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Crucial is the completeness of $Y$. Take any non-complete normed space $V$, choose $Y=V$ and $T$ the identity operator. Let $X$ be the completion of $V$ and assume $\overline{T}:X\to V$ the continuous extension of $T$.

As $V$ is not complete, there is a sequence $(v_n)_n$ in $V$ converging to some $x\in X\setminus V$. Then $\overline{T}(v_n)$ converges to $T(x)\in V$, but $T(v_n)=v_n$, so $x=T(x)\in V$, a contradiction.

If $Y$ is complete, such an extension always exists (see e.g. Theorem 5.1 in F. Treves' book "Topological Vector Spaces, Distributions and Kernels").

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  • $\begingroup$ You're invoking some definition of completeness for locally convex spaces here? $\endgroup$
    – Mike F
    Jan 22, 2014 at 22:53
  • $\begingroup$ Not for the counterexample, but for the existence theorem (see the mentioned book). $\endgroup$
    – Vobo
    Jan 22, 2014 at 22:55
  • $\begingroup$ @Mike Topological vector spaces are uniform spaces. A uniform space is complete (by definition) if every Cauchy filter/Cauchy net converges. $\endgroup$
    – Daniel Fischer
    Jan 22, 2014 at 22:58

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