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Let's denote by $T: X \to X$ a closed linear operator on an infinite dimensional Banach space. If needed, we can assume $X = L^p(\mathbb{R}^n)$ for some $p \neq 2, p > 1$. Furthermore $U_i, i \in \mathbb{N}$, are eigenspaces of $T$ for the eigenvalues $\lambda_i, i\in\mathbb{N}$.

Here comes my question: If $U$ denotes the closure of the direct sum $\oplus_{i\in\mathbb{N}} U_i$, is then $U$ invariant under $T$?

Of course, the first question already is, if $U$ is contained in the domain of $T$.

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    There's no reason why $U$ should be contained in the domain. Consider $X = \ell^p(\mathbb{N})$ and $T(e_n) = 2^n e_n$. – 2017-01-19
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    Thanks @DanielFischer. Now it's clear that this cannot hold true in such generality. Anybody an idea, if there are reasonable conditions ensuring invariance if the eigenspaces $U_i$ are finite dimensional but the eigenvectors do not build a Schauder basis as in Daniel's example? – 2017-01-19
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    What exactly is the "direct sum," here? Do you just mean the closed linear span? That is, do you mean $U=\overline{\text{span}}(\bigcup_{i\in\mathbb{N}}U_i)$ ? – 2017-01-21
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    It is obvious that if $\mathcal{V}$ is a family of $T$-invariant subspaces then $\text{span}\bigcup\mathcal{V}$ is also $T$-invariant. And if $T$ is continuous then so is $\overline{\text{span}}\bigcup\mathcal{V}$. But I do not think this last fact is true for discontinuous operators, even if $T$ is closed and $\overline{\text{span}}\bigcup\mathcal{V}\subseteq\text{dom}(T)$. – 2017-01-21
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    @BenWallis thanks. That's what I ment with direct sum. And I think so too now, that U usually won't be contained in the domain if T is closed and unbounded. – 2017-01-21
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    Well, it is still an interesting question under what conditions $\textbf{dom}(T)\cap\overline{\text{span}}\bigcup\mathcal{V}$ is $T$-invariant. – 2017-01-21

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