Let $\mathcal{H}_i$ be a collection of Hilbert spaces, and $T_i \in \mathbf{B}(\mathcal{H}_i)$. Assume that $\sup ||T_i|| = \infty$. Consider $\bigoplus \mathcal{H}_i$(the completetion). Given an operator $T$ defined on a subspace $D(T)$ into $\bigoplus \mathcal{H}_i$ where $D(T) := \{f \in \bigoplus \mathcal{H}_i: \sum ||T_i(f(i))||^2 < \infty \}$. Essentially this is the definition of an unbounded operator. Where $T := \bigoplus T_i$ i.e. $T(f)(i) := (T_i(f(i))$. I want to show that the graph of $T$ is closed i.e. $G(T) = \{(f, Tf): f \in D(T)\}$ is closed in $\bigoplus \mathcal{H}_i \oplus \bigoplus \mathcal{H}_i$. Given a limit point in this graph $(f, g)$, there is a sequence $(f_n, Tf_n)$ that converges to $(f, g)$. I have shown that $\lim f_n(i) = f(i)$. Since $T_i$ is bounded, then $\lim T(f_n(i)) = T(f(i))$. I am having trouble showing that $\sum ||T_i(f(i))||^2 < \infty$.
Also assume each $T_i$ is self adjoint(I don't know if that will help)