Linear operator between infinite dimensional vector spaces

Question

I found (in a book) an example of a linear operator $T$ between infinite dimensional vector spaces (in this case Hilbert spaces) $$T:H\rightarrow H',$$ whose image does not coincide with the entire space $H'$: $$ Te^{(n)}=e^{(n)}/n, $$ where $n=1,2,\dots$

The image of this operator does not include the vector $y=\sum e^{(n)}/n$.

I do not quite understand this, since $y$ seems like a linear combination of vectors of the image.

Answer 1

I suppose $(e^{(n)})$ stands for an orthonormal basis here. Since the basis is suppose to be contained in $H'$ for the definition of $T$ to make sense I will assume that $H'=H$.

Elements of $H$ are of the type $x=\sum\limits_{k=1}^{\infty} a_n e^{(n)}$ with $\sum\limits_{k=1}^{\infty} |a_n|^{2} <\infty$; also $T(x)=\sum\limits_{k=1}^{\infty} \frac {a_n} n e^{(n)}$. The element $y=\sum\limits_{k=1}^{\infty} \frac 1 n e^{(n)}$ belongs to $H$ because $\sum \frac 1 {n^{2}} <\infty$. But if $y=Tx$ the we get $a_n =1$ for all $n$ contradicting the fact that $\sum\limits_{k=1}^{\infty} |a_n|^{2} <\infty$.

Answer 2

No, it is not a linear combination. Linear combinations are finite sums.

Assuming that $H=H'=\ell^2$, yes, the statement is correc. The sequence $\left(1,\frac12,\frac13,\ldots\right)$, which is an ellement of $\ell^2$, does not belong to the range of $T$, since it could only be the image of $(1,1,1,\ldots)$, which is not an element of $\ell^2$.