Proof of uniqueness of the bounded linear transformation extended in the Bounded Linear Transformation theorem

B.L.T Theorem (from Reed/Simon): Suppose $T$ is a bounded linear transformation from a normed linear space $\langle V_1, \|\cdot\|\rangle$ to a complete normed linear space $\langle V_2, \|\cdot\|\rangle$. Then $T$ can be uniquely extended to a bounded linear transformation (with the same bound), $\widetilde{T}$, from the completion of $V_1$ to $\langle V_2, \|\cdot\|\rangle$.

The proof for $\widetilde{T}$ being bounded was given and very straightforward, and proving that it was linear was pretty simple as well. I have been having issues with proving that $\widetilde{T}$ is unique, however, despite it probably being easy. I tried supposing towards a contradiction and using that $V_1$ is dense in its completion, $\tilde{V_1}$, the extending transformations must agree on $V_1$, the extending transformations must both have the same bound as $T$, and since this implies that the extending transformations must have different bounds to be different themselves, this proves that $\widetilde{T}$ is unique. I don't think that this reasoning is right since I think that a transformation can act differently on some subset of $\tilde{V_1}\setminus{V_1}$ without changing the bound, and my only other idea was to use that both of these spaces are $T_1$, so if a sequence converges, it must converge to a unique point, and since $T$ is bounded and therefore continuous, and any extensions must be bounded and therefore continuous, we'll have sequences being mapped to their limits, and since these limits are unique, we will only yield one extension $\widetilde{T}$ that works for all of $\tilde{V_1}$. Again, I think that this reasoning is missing something.

Any insights, whether it be a nudge in the right direction or full proofs, are very welcome! Thanks in advance.

Let $y\in \bar{V_1}$, then there exists a sequence $\{x_n\}_n\in V_1$ such that $x_n\rightarrow y$. Now consider$\|T(x_m)-T(x_n)\|$,

then we have $\|T(x_m)-T(x_n)\|=\|T(x_m-x_n)\|\leq C\|x_m-x_n)\|$ (because $T$ is bounded linear). Since convergent sequences are cauchy, hence for large $m, n$ implies $\|x_m-x_n)\|$ goes to zero. That implies $\{T(x_n)\}_n$ is cauchy in $V_2$ and $V_2$ is complete, limit of $\{T(x_n)\}_n$ exists.

Define $\bar{T}(y)=\lim T(x_n)$. Next we want to show $\bar{T}$ is well-defined. Let $\{x'_n\}$ be sequence such that $x'_n\rightarrow y$. Then $\|T(x_n')-T(x_n)\|\leq C\|x_n'-x_n)\|\rightarrow 0$ as $n\rightarrow \infty$. This implies $\lim T(x_n')=\lim T(x_n)=\bar{T}(y)$. hence $\bar{T}$ is well-defined.

Next we want to show $\bar{T}$ is linear. Take $x, y\in \bar{V_1}$ and $\alpha, \beta\in\mathbb{R}$, then there exists two sequences $(x_n)_n$ and $(y_n)_n$ such that $x_n\rightarrow x$ and $y_n\rightarrow y$ as $n\rightarrow \infty$. Clearly $\alpha x_n+\beta y_n\rightarrow \alpha x+\beta y$. Then,

$\bar{T}(\alpha x+\beta y)=\lim T(\alpha x_n+\beta y_n)=\alpha \lim T(x_n)+\beta \lim T(y_n)=\alpha\bar{T}(x)+\beta\bar{T}(y)$. Therefore $\bar{T}$ is linear. Next need to show $\bar{T}$ is bounded. For any $y\in\bar{V_1}$, we have

$\bar{T}(y)=\lim T(x_n)\leq \lim C\|x_n\|= C\lim\|x_n\|=C\|y\|$, and hence $\bar{T}$ is bounded. Next we need to check what will happen when we choose $y\in V_1$ (not in closure of $V_1$). Choose constant sequence $(y_n)=y$ for all $n$. Then

$\bar{T}(y)=\lim T(y_n)=T(\lim y_n)=T(y)$. Therefore, $T$ can extend to closure of $V_1$.

Finally, we want to show uniqueness of $\bar{T}$. Suppose there is an another extension $B$. For any $y\in\bar{V_1}$, there exists a sequence $(x_n)$ such that $\lim x_n=y$. Then,

$\bar{T}(y)=\lim T(x_n)=\lim B(x_n)=B(\lim x_n)=B(y)$, hence extension is unique. This complete the proof.

Let $\{x_n\}$ be a Cauchy-Sequence in $V_1$. Then $\{T(x_n)\}$ is a Cauchy-Sequence in $V_2$. (This follows from the boundedness of $T$)

Then the natural way to extend $T$ is using the limit of Cauchy-Sequence in $V_2$. (which is possible by completeness of $V_2$)

For the proof of uniqueness, suppose there are two extensions $T_1$ and $T_2$, and use the fact that $T_1-T_2$ is bounded linear. (hence continuous)

Apply to the Cauchy-Sequence $\{x_n\}$. Then for $x^*=\lim x_n \in \bar{V_1}$, $(T_1-T_2)(x_n) = 0$ for all $n$. Thus $T_1(x^*) = T_2(x^*)$ by continuity.