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I am studying unbounded operators and the graphs of those operators. I found that the closure of a graph may not be the graph of any operator. Can someone provide an example of an operator and a graph, $\Gamma(T)$, for which $\overline{ \Gamma(T)}$ is not the graph of an operator?

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  • $\begingroup$ These are what we call ‘non-closable operators’. $\endgroup$ – Berrick Caleb Fillmore Apr 25 '15 at 2:09
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Let $X$ be a Hilbert space with orthonormal subset $\{ e_{n} \}_{n=1}^{\infty}$. Define $F : \mathcal{D}(F) \subset X \rightarrow \mathbb{C}$ by $F(x) = \sum_{n=1}^{\infty}n(x,e_n)$ on the domain $$ \mathcal{D}(F)=\{ x \in X : \sum_{n=1}^{\infty}|n(x,e_n)| < \infty \}. $$ Then $\{ \frac{1}{n}e_{n} \}_{n=1}^{\infty}$ converges to $0$ and $F(\frac{1}{n}e_{n})=1$ converges to $1$. Therefore $(0,1)$ is in the closure of the graph of $F$. But $(0,1)$ is not in the graph of any linear $G : \mathcal{D}(G)\subseteq X \rightarrow \mathbb{C}$.

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