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Let $\mathcal{H}$ be a Hilbert space. Suppose that $T,S$ are closed range operators such that the subspaces $T\mathcal{H}$ and $S\mathcal{H}$ are orthogonal. Consider the operator $T+S$. Is the sum operator closed range?

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1 Answer 1

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Consider the operators on the Hilbert space $\ell^2$ defined as follows

$$\eqalign{(S x)_{2n} &= x_{2n} \cr (S x)_{2n+1} &= 0 \cr (Tx)_{2n} &= 0 \cr (T x)_{2n+1} &= x_{2n} + \frac{x_{2n+1}}n \cr} $$ It is easy to check that $S$ and $T$ both have closed ranges, which are orthogonal. Indeed, $S \ell^2 = \{x \in \ell^2: x_{2n+1} = 0 \text{ for all }n\}$ and $T \ell^2 = \{x \in \ell^2: x_{2n} = 0 \text{ for all }n\}$. But $S+T$ does not have closed range. Note that $(S+T) \ell^2$ includes all vectors of finite support (so it is dense), but does not contain $y$ with all $y_{2n} = 0$ and all $y_{2n+1} = 1/n$ (as $(T+S) x = y$ would imply $x_{2n} = 0$ and $x_{2n+1} = 1$).

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