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Given the Banach space $X:=\mathcal{C}([0,1]\cup[2,3])$.

I remember I've seen a beautiful example of a non-closable operator whose graph is dense.
It involved exploiting Stone-Weierstraß for a disconnected domain and playing around with polynomials for defining the operator but I simply can't puzzle together the example anymore.
Do you by any chance know about such a construction or help me figuring it out by hand?

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Let $\mathcal D$ be the subspace of $X$ consisting of functions of the form $f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]}$ where $f_1$ and $f_2$ are polynomials and $\chi_I$ is the indicator function of $I$. On this domain consider the operator $T$ defined by $T(f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]}) = f_2 \chi_{[0,1]} + f_1 \chi_{[2,3]}$. To see that this has dense graph, note that for any $g_1, h_1 \in C[0,1]$ and $g_2, h_2 \in C[2,3]$ there are polynomials $f_1, f_2$ such that $f_1$ approximates $g_1 \chi_{[0,1]} + h_2 \chi_{[2,3]}$ and $f_2$ approximates $h_1 \chi_{[0,1]} + g_2 \chi_{[2,3]}$. Then $f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]}$ approximates $g_1 \chi_{[0,1]} + g_2 \chi_{[2,3]}$ while $T(f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]})$ approximates $h_1 \chi_{[0,1]} + h_2 \chi_{[2,3]}$.

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  • $\begingroup$ Aha, and we know that polynomials extend uniquely, i.e. $f\chi_I=f'\chi_I\implies f=f'$. Thank you very much! :) $\endgroup$ – C-Star-W-Star Jun 3 '16 at 16:15

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