# Bounded linear operator - translation operator.

The image below is from my lecture notes for Linear analysis:

Shouldn't it be $T:X\to X$? If not, I have no idea what it means, because if $z$ is in $X$ then there is no reason as to why $x + z$ should be in $Y$.

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Could you write it in TeX? I don't get what is is written there! – Norbert May 7 '12 at 22:50
@Norbert "Because $T$ is linear, you can use translations. For $x\in X$, the map \eqalign{X&\rightarrow Y \cr z&\mapsto x+z } is continuous..." – David Mitra May 7 '12 at 22:54
Yeah that's what it looks like. I think it is a mistake and the easiest way to correct is it to change Y to X. The only other thing I think it could mean is: get a linear operator T, and y in Y, and map x (in X) to T(x) + y. Not sure this works. Anyhow, I think the first way was intended. – Adam Rubinson May 7 '12 at 22:58
@DavidMitra, thanks! – Norbert May 7 '12 at 23:08
@AdamRubinson I can see that after the definition of that map its inverse is also mentioned. From their forms it is clear that there is no need in $T$ operator. – Norbert May 7 '12 at 23:10