An exercise on a closed graph, functional analysis I'm studying functional analysis for the first time, if possible I would need a review of the following exercise, thanks in advance.
Let $(E,\| \cdot \|_{E})$ and $(F,\| \cdot \|_{F})$ be Banach spaces. Let $T: E \mapsto F$ be a linear operator and denote by $G(T)$ its graph, that is $G(T)=\{(x,Tx) : x \in E \}$. Prove that $G(T)$ is closed if and only if $(E,\| \cdot \|_{1})$ is a Banach space, where $\|x\|_{1}=\|x\|_{E}+\|Tx\|_{F}$.
My solution:
"$\Longrightarrow$"
The hypothesis of the closed graph theorem are satisfied, hence $T$ is continuous (and bounded). For every $x \in E$ it holds:
$$ \|x\|_{E} \le \|x\|_{1} = \|x\|_{E}+\|Tx\|_{F} \le \|x\|_{E}+\| T \|_{\mathcal{L}(E,F)} \| x \|_{E} = \|x \|_{E} (1+\| T \|_{\mathcal{L}(E,F)}) $$
It follows that the two norms are equivalent on $E$. Being $(E,\| \cdot \|_{E})$ complete, we conclude that  $(E,\| \cdot \|_{1})$ is a Banach space.
"$\Longleftarrow$"
Suppose $(E,\| \cdot \|_{1})$ is complete. Let $((x_n,Tx_n))_n \subset G(T)$ be a sequence converging to a limit point in $E \times F$ (with respect to the norm $\| \cdot \|_{E \times F}=\| \cdot \|_{E}+ \| \cdot \|_{F}$), say $(x,y)$. Since 
$$ \|x_n-x\|_{E}+\|Tx_n-y\|_{F}=\|(x_n,Tx_n)-(x,y)\|_{E \times F} \rightarrow 0,$$
it follows that $Tx_n \rightarrow y $ in $(F,\| \cdot \|_{F})$ and $x_n \rightarrow x $ in $(E,\| \cdot \|_{E})$; in particular, these sequences are Cauchy's in $E$ and $F$ respectively.
Then $(x_n)_n$ turns out to be Cauchy's with respect to $\| \cdot \|_{1}$ too, in fact for every $m,n \in \mathbb{N}$:
$$ \| x_m - x_n \|_{1} =\| x_m - x_n\|_{E} + \| T(x_m-x_n) \|_{F} = \| x_m - x_n\|_{E} + \| Tx_m - Tx_n \|_{F}  $$
Since $(E, \| \cdot \|_1 )$ is complete, $(x_n)_n$ must converge to a limit $\overline{x}$ with respect to $\| \cdot \|_1$:
$$ \|x_n-\overline{x} \|_{E} + \|Tx_n - T\overline{x} \|_{F} = \| x_n-\overline{x} \|_1 \rightarrow 0,$$
whence $x_n \rightarrow \overline{x}$ in $(E,\| \cdot \|_{E})$ and $Tx_n \rightarrow T\overline{x}$ in $(F, \| \cdot \|_{F}$). By the uniqueness of the limit we get $x=\overline{x}$ and $Tx=y$, so that $(x,y) \in G(T)$.
 A: Your reasoning is sound.  If forced to make a suggestion, I would say that the proof of "$\Longleftarrow$" includes some unnecessary details, namely the result that $x_n\to x$ and $Tx_n\to y$ (in the original norms of $E$ and $F$).  You could shorten the proof as follows.

Suppose $(E,\|\cdot\|_1)$ is complete. Let $\{(x_n,Tx_n)\}$ be a sequence in $G(T)$ convergent to $(x,y)$ in $E\times F$.  (Recall that the product topology on $E\times F$ is generated by the norm $\|(x,y)\|_{E\times F}=\|x\|_E+\|y\|_F$.) In particular $\{(x_n,Tx_n)\}$ is Cauchy, and thus $\{x_n\}$ is Cauchy in $(E,\|\cdot\|_1)$, hence convergent to some $x'\in E$ with respect to $\|\cdot\|_1$.  Furthermore, for any $n$, we have
  \begin{align}
 \|(x',Tx')-(x,y)\|_{E\times F}
&\leq \|(x',Tx')-(x_n,Tx_n)\|_{E\times F}+\|(x,y)-(x_n,Tx_n)\|_{E\times F} \\
&=\|x'-x_n\|_1+\|(x,y)-(x_n,Tx_n)\|_{E\times F}
\end{align}
  Thus $x=x'$ and $y=Tx$, so the closed graph theorem applies and therefore $T$ is continuous.

Of course, this is just a cosmetic difference. Good work.
