Jtdylktuy

I am working through Hueser's Functional Analysis and have come across the proof (7.4) that:

If E is a normed space and F is a Banach space then $L(E,F)$, the set of continuous linear transformations from E to F is a Banach space.

The proof constructs a Cauchy sequence of operators and uses the fact that F is a Banach space to arrive at the result:

begin{equation} ||A_nx-Ax|| < epsilon||x|| end{equation}

Where $A_nx rightarrow Ax$ in F and ${A_j}$ constitutes a Cauchy sequence of operators.

My confusion arises as to how this implies $A_n rightarrow A$ as the author's just arrives at this result in the next line with no explanation.

My idea is that because $Ax$ is linear and continuous and E is a Vector space so contains the identity therefore set $x=1$ and get the result.

However any clarification or hints would be appreciated.

To show that $A_n to A$ in $L(E,F)$, we need to show that $||A_n - A|| to 0$, which is equivalent to saying that for any $epsilon > 0$, there exists $N$ such that if $n > N$ , we have $||A_n - A|| < epsilon$.

However, $||A_n - A|| < epsilon$ is the same as $||(A_n - A)x|| < epsilon ||x||$ for all $x in E$ , which is the same as $||A_n x - Ax|| < epsilon ||x||$ for all $x in E$, which is the statement you have.

In other words, the last statement is the assertion that $||A_n - A|| < epsilon$, which was to be shown for showing $A_n to A$. This equivalence was omitted in the text since it nearly follows by definition.