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Let $X$ be a Banach space and $T$ a linear bounded operator defined on $L(X,Y)$ with $Y$ a normed space. If $T$ is an isometry then $TX$ is a closed subspace of $Y$.

I considered a sequence $y_n$ in $TX$. I want to show that the limit of this sequence is in $TX$. $y_n \in TX$ it means it exists $x_n \in X$ such that $Tx_n=y_n$. Being that $T$ is an isometry it is necessary a one-one map so it exists the inverse operator $T^{-1}$. So $x_n=T^{-1}y_n$. Making the limit $x=T^{-1}y$ but $x_n$ tends to $x$ which belongs to $X$ because $X$ is a Banach space. So $y=Tx \in TX$ and the thesis follows. Is my proof correct?

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You never explicitely use the isometry. The claim is in general false if $T$ is only assumed injective. You should use that with $(y_n)_n$ also $(x_n)_n$ is Cauchy (why?). – PhoemueX May 27 '14 at 8:49
If I take $y_n$ to be Cauchy then $||y_n-y_m||=||Tx_n-Tx_m||=||T(x_n-x_m)||=||x_n-x_m||$ because of the isometry. Sending $m$ to infinite so we have $||x_n-x||=||Tx_n-Tx||=||y_n-y||$. Being $X$ Banach $x_n$ converge to $x \in X$ and so does $y_n \in Y$. I don't know I feel like I still miss some points... – user73793 May 27 '14 at 9:35
First, $(y_n)$ is automatically Cauchy, because you take a convergent(!) sequence $y_n \rightarrow y$ from $TX$ with $y \in Y$ and want to show that $y \in TX$ as well. As you show, the $(x_n)_n$ are then also Cauchy, thus convergent to some $x \in X$ (important for this: $X$ is complete). The other part of your comment implies $y_n = Tx_n \rightarrow Tx$ and thus $y = Tx \in TX$. – PhoemueX May 27 '14 at 9:45
Yes understood what you mean. Thanks. – user73793 May 27 '14 at 9:48
up vote 0 down vote accepted

As PhoemueX said, the proof lacks the consideration of the continuity of $T^{-1}$. Once that is added, the proof is correct.

But I would put the proof differently:

  1. The image of a complete metric space under a distance-preserving map is also a complete metric space.

  2. If a subset of a metric space is complete in the induced metric, then that subset is closed.

My point here is that linearity has little to do with the problem. It's mostly the metric.

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