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Consider $X$ a normed vector space, $Y$ a Banach space and $T:D_{T} \rightarrow Y$ a linear map such that $D_{T}$ is dense in $X$ and $T$ is continuous on $D_{T}$.

I have to prove that $T$ admits a unique continuous exstension $T: X \rightarrow Y$; that is there is a unique $T\in \mathcal{L}(X,Y)$ such that $T_{|D_{T}}=T$.

This is my idea:

Let $y \in X$. Since $D_{T}$ is dense in $X$ exists a sequence $\{x_n \}$ in $D_{T}$ such that $x_n \rightarrow y$. Since an operator between normed spaces is continuous iff it is bounded I have $\forall n,m$: \begin{equation} ||Tx_m-Tx_n||=||T(x_m-x_n)||\leq C||x_m-x_n|| \end{equation} Convergent sequences are cauchy, hence for large $n,m$ implies $||x_m-x_n||$ goes to zero. That implies $\{Tx_n\}$ is cauchy in $Y$. Since $Y$ is complete, limit of $\{Tx_n\}$ exists. So I prove that \begin{equation} \lim_{y\rightarrow x, y \in D_T } Ty \end{equation} converge. Now, I extend $T$ defining $Tx$ as the value of the limit.

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  • $\begingroup$ I'm sorry. I put right immediatly $\endgroup$ – JohnMalkovich Jan 12 '18 at 16:40
  • $\begingroup$ (1) do you know the fact that an operator between normed spaces is continuous iff it is bounded? (2) The continuity assumption only gives you that if $x_n$ converges to a point $x$ of $D_T$, then $T x_n \to Tx$. Here $x$ need not be in $D_T$, so you have to work harder. $\endgroup$ – Nate Eldredge Jan 12 '18 at 17:03
  • $\begingroup$ "I suppose that an extension of $T$ exists". No - you only do that if you are trying to disprove the statement by contradiction. The only thing you can prove starting this way is that "if $T$ exists, then it exists", which is obvious. Instead define a function $T$ by (1)showing that $$\lim_{y \to x, y \in D_T} Ty$$ must converge (2) Defining $Tx$ to be the value (3) showing that $T$ is linear and continuous. $\endgroup$ – Paul Sinclair Jan 12 '18 at 19:35
  • $\begingroup$ @PaulSinclair I've modified my idea for the proof, I hope it is better than the previous one $\endgroup$ – JohnMalkovich Jan 13 '18 at 16:24
  • $\begingroup$ Definitely better. It would be more general if you didn't rely on the existence of sequences, but the such sequences are guaranteed to exist in a Banach space, so you are okay there (except you need to say $\lim_n$ instead of copying the more general limit I mentioned, as you haven't shown the general version, and also I used $y$ differently than you have, so you are using it twice to mean different things now, which is bad form). Now you have to prove that your extended $T$ is still (1) linear (2) continuous, and that it is unique. $\endgroup$ – Paul Sinclair Jan 13 '18 at 16:55

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