If $z$ is a fixed element of a normed space $X$ and $f \in X'$, show that $T: X \rightarrow X$ defined by $Tx=f(x)z$ is compact.

Question

If $z$ is a fixed element of a normed space $X$ and $f \in X'$, show that $T: X \rightarrow X$ defined by $Tx=f(x)z$ is compact.

The solution I have says that the range is at most one dimensional and so compactness follows.

I understand why the dimension being finite implies compactness, what I don't understand is how we know the dimension of the range is at most $1$.

I feel like I am missing something basic here, but my thought is that we don't know the dimension of $X$, so how do we know the dimension of $Tx$?

Answer 1

Because all elements of the range of $T$ are muliples of $z$ and the space $\{\lambda z\,|\,\lambda\text{ is a scalar}\}$ is $1$-dimensional.

Answer 2

This is because $f(x)z\in\textrm{span}(z)$ for any $x\in X$, since $f\in X'$ implies $f(x)$ is real (complex) number.