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Let $X,Y$ be Banach spaces and $T:D_T \subset X\to Y$ a linear (not necessarily bounded) operator. Let $D_{T,\mathrm{graph}}$ denote $D_T$ endowed with the norm given by $$ \|x\|_\mathrm{graph}:=\|x\|+\|Tx\|. $$ Show that if $T$ is a closed operator, then $D_{T,\mathrm{graph}}$ is a Banach space and $T\in \mathcal{B}(X,Y)$. Show also that the norm of $T$ in $\mathcal{B}(D_{T,\mathrm{graph}},Y)$ is $1 $ if and only if $T$ is unbounded as an operator from $D_T$ (with the $X$-norm) to $Y$.

I can solve the first part. But I don't know how to show the last sentence.

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Hint: compare the inequalities $\|Tx\|\le A\|x\|$ and $\|Tx\|\le B(\|x\|+\|Tx\|)$. What implies what here? –  user53153 Feb 20 '13 at 17:22
    
Should I discuss different cases when $A=1$, $A>1$ and $A<1$? –  user60610 Feb 20 '13 at 17:30
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No. I'll make my hint more precise: relate $\dfrac{\|Tx\|}{\|x\|}$ to $\dfrac{\|Tx\|}{\|x\|+\|Tx\|}$ algebraically. Then explore the implication for the supremum of such things (which is the operator norm). –  user53153 Feb 20 '13 at 17:32
    
$\frac{\|Tx\|}{\|x\|+\|Tx\|}=\frac{1}{\frac{\|x\|}{\|Tx\|}+1}$? Am I thinking on the right track? –  user60610 Feb 20 '13 at 18:00
    
Yes. Now consider what happens when the supremum of the fraction on the left is equal to $1$. –  user53153 Feb 20 '13 at 18:04

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Thanks to @5pm's help, this problem is solved. $$ \sup \frac{\|Tx\|}{\|x\|+\|Tx\|}=\frac{1}{\inf\frac{\|x\|}{\|Tx\|}+1}=\frac{1}{\frac{1}{\sup\frac{\|Tx\|}{\|x\|}}+1} $$ This is equivalent to $$ \|T\|_{graph}=\frac{1}{\|T\|^{-1}+1} $$ This equation establishes the last claim in the problem.

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