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!
 A: 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.


*

*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 $x_n \to x$.

*Check that $\tilde{T}|_A = T$ (consider constant sequences).

*Check that $\tilde{T}$ is linear (obvious from basic properties of limits).

*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 $\tilde{I}: c_0 \to c_{00}$.
