In a normed space, the sum of a Closed Operator and a Bounded Operator is a Closed Operator.

Question

The book "Introductory Functional Analysis with Applications" (Kreyszig) presents the following lemma

Let $T:\mathcal{D}(T)\to Y$ be a bounded linear operator with domain $\mathcal{D}(T)\subset X$, where $X$ and $Y$ are normed spaces. Then:

(a) If $\mathcal{D}(T)$ is a closed subset of $X$, then $T$ is closed.

(b) If $T$ is closed and $Y$ is complete, then $\mathcal{D}(T)$ is a closed subset of $X$. (page 295)

and the following problem

Let $X$ and $Y$ be normed spaces. If $T_1:X\to Y$ is a closed linear operator and $T_2\in\mathcal{B}(X,Y)$, show that $T_1 + T_2$ is a closed linear operator. (page 296)

If $X$ and $Y$ are Banach spaces, then we can conclude (by Closed Graph Theorem) that $T_1$ is bounded. So, $(T_1+T_2):X\to Y$ is bounded and thus it's closed (by lemma above).

My question is: how to solve it when $X$ and $Y$ are not necessarily complete spaces?

Thanks.

Answer 1

Possible Method: Let $x_n \to x \in X$ and $(T_1+T_2)x_n \to y \in Y$. What you need to show is that $(T_1+T_2)x = y$. Now, $T_2 x_n \to T_2 x$ since $T_2 : X \to Y$ is continuous. From here we can prove that $T_1 x_n \to y-T_2 x$. Indeed $|T_1 x_n + T_2 x - y| \leq |T_1 x_n + T_2 x_n - y| + |T_2 x_n - T_2 x| \to 0$. Therefore $T_2 x_n \to y - T_1 x$. Since $T_1$ is closed this means that $y-T_2 x = T_1 x$. Equivalently $y = (T_1 + T_2)x$.