# If $D$ is a dense linear subspace of $X$ then $D\to Y$ extends to $X\to Y$ uniquely

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.

• Right, bounded linear operators are uniformly continuous (Lipschitz continuous even). – Daniel Fischer May 18 '15 at 23:14
• 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. – YTS May 18 '15 at 23:15

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., $\|\tilde{T}\| = \|T\|$.

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