# Extensions of Bounded Linear Operators

I am currently lacking in some basic knowledge regarding extensions of linear operators. Let $X$ and $Y$ be Banach spaces and let $A$ be dense subspace of $X$. Let $T : A \to Y$ be a bounded linear operator.

Is it true that there exists a unique extension $\tilde{T} : X \to Y$ of $T$ that is also bounded?

I believe this is common result but I can't seem to figure out what to search if there is a name for this result, or is this somehow a consequence of the Hahn-Banach theorem?

Also is this result true if $X$ and $Y$ are just normed spaces?

Any help would be appreciated. Thanks!

The proof is a little bit long to write down so let me just sketch the main ideas here.

The main thing to notice is that there is really only one way to define $\tilde{T}$ as a continuous extension of $T$. Since $A$ is dense in $X$, for every $x \in X$ there is a sequence $x_n \in A$ such that $x_n \to x$. We want $\tilde{T}$ to be a continuous map such that $\tilde{T}|_A = T$ so we must have $$\tilde{T}x = \lim_{n \to \infty} Tx_n.$$

So the thing to do is to make the above equation your definition of $\tilde{T}$ and check that it works. To do this you need to do a few things.

1. Check that $\tilde{T}$ is well-defined. This requires us to check that the limit in the definition really exists (here you will use that $Y$ is complete) and that the limit is independent of the choice of sequence $x_n \to x$.
2. Check that $\tilde{T}|_A = T$ (consider constant sequences).
3. Check that $\tilde{T}$ is linear (obvious from basic properties of limits).
4. Check that $\tilde{T}$ is bounded. For this notice that $\|\tilde{T}x\| = \lim_{n \to \infty} \|Tx_n\| \leq \lim_{n \to \infty} \|T\| \cdot \|x_n\|$.

In proving these four points, you should only ever need to assume the completeness of $Y$. All that remains is to see that completeness of $Y$ is in fact a necessary condition.

To do this consider $c_{00}$ as a dense subspace of $c_0$ (both with the $\infty$-norm). Then the identity map $I:c_{00} \to c_{00}$ has no continuous extension $\tilde{I}: c_0 \to c_{00}$.

• Ah, I see it now! Thank you for the great response! – LMW Jul 2 '18 at 18:37
• Dear Professor Rhys. Could you please provide me a reference which contains the proof of the extension theorem because I want to cite it. Thanks a lot. – Student Mar 1 '19 at 13:11
• @Student Try Theorem 2.7-11 in Introductory Functional Analysis by Kreyszig. Btw I'm not a professor, it's fine to just refer to me by name :) – Rhys Steele Mar 1 '19 at 15:33
• @Teodorism I don't use Hahn-Banach at all in this answer. In fact, this result is much more elementary than the Hahn-Banach. – Rhys Steele Dec 8 '19 at 18:33
• @Teodorism $T$ is the operator defined on $A$ that you started with. This operator is assumed to be bounded. – Rhys Steele Dec 8 '19 at 22:33