3
$\begingroup$

I am trying to prove the following, but I am not confident in my work.

Let $D$ be a linear subspace of a normed space $X$ that is dense in $X$. Let $Y$ be a Banach space. Show that any bounded linear operator $T : D \to Y$ has a unique extension to a bounded linear operator $X \to Y.$

What I tried. I can prove the lemma

Lemma: Assume that $D$ is a dense subset of $X_1$, that $(X_2, d_2)$ is complete and that $f : D → M_2$ is a uniformly continuous mapping. Then there exists a unique continuous mapping $F : M_1 → M_2$ such that $F|_D = f$.

Bounded linear operators are uniformly continuous the existence and uniqueness of a continuous extension follows from the Lemma. If I am correct, I can use the proof of the lemma to construct the desired extension.

$\endgroup$
  • $\begingroup$ Right, bounded linear operators are uniformly continuous (Lipschitz continuous even). $\endgroup$ – Daniel Fischer May 18 '15 at 23:14
  • $\begingroup$ I think you are right. By continuity you can extend the function to all $X_1$. You should only prove that it is indeed bounded. $\endgroup$ – YTS May 18 '15 at 23:15
5
$\begingroup$

What you said is correct. The following proof is essentially the same proof as you will give for your lemma

HINT: For any $x\in X$ let $(x_n)$ be a sequence in $D$ such that $x_n\longrightarrow x$. Since $T$ is linear and bounded, we have \begin{equation} \|Tx_n-Tx_m\| = \|T(x_n-x_m)\|\leq\|T\|\|x_n-x_m\|. \end{equation} This shows that $(Tx_n)$ is a Cauchy sequence because $(x_n)$ converges. By assumption that $Y$ is a Banach space $(Tx_n)$ converges, say, \begin{equation} Tx_n\longrightarrow y\in Y. \end{equation} Define $\tilde{T}$ by \begin{equation} \tilde{T}x=y. \end{equation} Now check the following:

(1) $\tilde{T}$ is well defined.

(2) $\tilde{T}$ is linear and is an extension of $T$.

(3) $\tilde{T}$ is bounded. To see this notice that $\|Tx_n\|\leq \|T\|\|x_n\|$ and $x\mapsto \|x\|$ is a continuous mapping.

NOTE: By (2) you will get that $\|\tilde{T}\| \leq \|T\|$ and obviously $\|\tilde{T}\|\geq\|T\|$ because the norm, being defined by supremum, cannot decrease in an extension. So, in fact you have a norm preserving extension, i.e., $\|\tilde{T}\| = \|T\|$.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ (0) one needs to check that the limit $y$ does not depend on the choice of the sequence. $\endgroup$ – daw May 19 '15 at 6:23
  • $\begingroup$ @daw Thank you very much for pointing this out........ $\endgroup$ – Urban PENDU May 19 '15 at 12:20
  • $\begingroup$ Not really important, but the note should say "By (3)" rather than "By (2)". $\endgroup$ – A. Howells Aug 8 '18 at 15:18

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.