# A linear operator $Tx=(x_1,x_1+x_2,x_1+x_3,\dots)$ between $c_0$ and $c$ and its norm

The operator $T$ is for $x=(x_n)_{n=1}^\infty \in c_0$ given by $$Tx=(x_1,x_1+x_2,x_1+x_3,\dots).$$ Show that $T$ is a bounded linear operator between $c_0$ and $c$ and calculate $||T||$.

Is $T$ injective, surjective?

First I would like you to have a look at my answer to the second question.

It is indeed injective as for each two distinct $x$ and $y$ the images are also distinct: if $x_1 \ne y_1$ (possibly some other coordinates are different too), then $Tx$ surely differs from $Ty$ in the first coordinate and if $x_1=y_1$ and there are differences between the two in some elements $n \ge2$, than $n$th coordinates of $Tx$ and $Ty$ are not equal.

As for surjectivity, I reckon $\lim_{n \rightarrow \infty} (x_1+x_n)=x_1+\lim_{n \rightarrow \infty}x_n=x_1$ and since $x_1$ is an arbitrary number then every number can be obtained as a limit but I feel I'm missing something here.

As far as $||T||$ goes, firstly let's take $x=(1,1,0,0,0,\dots) \in c_0, ||x||=1$ and the definition $||T||=\sup_{||x||=1}||Tx||$ yields $||T|| \ge 2$. On the other hand, $||Tx||=\sup_{n\ge2}\{|x_1|,|x_1+x_n|\} \le 2\sup_n|x_n|=2||x||$, hence $||T|| \le 2$. All in all, $||T||=2$.

I'd appreciate any remarks on the correctness of my solution.

• What are $c$ and $c_0$? Commented Jul 4, 2017 at 17:32
• @TZakrevskiy $c$ are convergent sequences and $c_0$ are sequences convergent to $0$. Commented Jul 4, 2017 at 17:45
• This is correct. Commented Jul 4, 2017 at 17:50
• @Aweygan Do you also think that $T$ is surjective? Does it imply that every sequence convergent to some $x_1$ can be represented as $(x_1,x_1+x_2,x_1+x_3,\dots)$? Commented Jul 4, 2017 at 17:57
• @Theta No, it is not surjective. The image contains only sequences with the first element being equal to the limit. For example, $(1,0,0,0,...)$ is not in the form $Tx$ for some $x\in c_0$.
– A.Γ.
Commented Jul 4, 2017 at 18:02

• For injectivity: what you did seems correct. Maybe this can be shortened a little bit, using linearity of $T$. Indeed, if $Tx=0$, then $x_1=0$ and for $n\geqslant 2$, $0=x_1+x_n=x_n$ hence $x=0$.
• For surjectivity: as pointed out by A.Γ., the range of $T$ is contained in the set of convergent sequences whose limit is equal to the first term, hence it is not the whole $c$. It is sufficient in order to prove that $T$ is not surjective, but finding the range could be also interesting.