# Proving that every bounded linear operator between normed linear spaces is closed?

I am working my way through a book on functional analysis and I'm unsure how to approach this problem:

Suppose $A:E_1 \to E_2$ is a bounded linear operator between two normed linear spaces $E_1$ and $E_2$. Show $A$ is closed in the sense that if the following hold:

1. $x_n \in D_A$
2. $x_n \to x$
3. $Ax_n \to y$

Then it must be true that

1. $x \in D_A$
2. $Ax = y$

My attempt:

Since the operator is between two normed linear spaces, then it follows that the operator is continuous since it is bounded (as boundedness and continuity is equivalent in spaces that satisfy the first axiom of countability).

I am unsure however of how far continuity can take me here. If $x \in D_A$ then continuity should imply (5), but how do I show that $x$ is necessarily in the domain of $A$?

For part 4:

Since you aren't referencing any ambient space(that contains $E_1$), I would assume that convergence in condition 2 is understood with respect to the norm on $E_1$. This would imply $x\in E_1$ because $E_1$ is closed in its own norm topology. But since $E_1=D_A$, this seems to imply 4. If you are referencing an ambient normed space $X$ though, where $E_1$ is a subspace that inherits the norm from $X$, I believe it is enough for $E_1$ to be closed.

For part 5:

Notice that since $A$ is bounded

$\|Ax_n-Ax\|=\|A(x_n-x)\|$

$\leq \|A\|\|x_n-x\|\to 0$

So that $Ax_n$ converges to $Ax$.

By uniqueness of limits in normed spaces(in particular, Hausdorff spaces), it follows that if $Ax_n\to y$

then $y=Ax$