What would be an example of an operator $T$ defined on a densely defined domain (subspace of banach space) such that domain of $T^*$ is $\{0\}$? In the text I am reading, I am presented of unbounded operators with adjoint having the same domain. For example, Let $X$ be a closed subset of $\mathbb{C}$ with the positive lebesgue measure and consider $L^2(X)$. Define $T$ on the domain $D(T) := \{f \in \mathcal{L}^2(X): \int |\lambda f(\lambda)|^2 d \lambda < \infty\}$ where $(Tf)(\lambda) := \lambda f(\lambda)$ where $\lambda \in X$. Then we have $D(T) = D(T^*)$.
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1If you have a closable operator on a Hilbert space, then the adjoint is always densely defined, so the domain of $T^\ast$ being ${0}$ is impossible. If you don’t require the operator to be closable, an example can be given. See my answer here. – David Gao Dec 02 '24 at 00:42
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2Also see here and here. – Dean Miller Dec 02 '24 at 01:22