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Let $H$ be a Hilbert space.

I know a lot of examples of linear operator $T$ on $H$ such that $D(T) \subsetneq H$ and $T$ is not bounded, and such kind of operators are very important in analysis of PDEs.

However, I don't know any example of linear operator $T$ on $H$ such that $D(T)=H$ and $T$ is not bounded. Does anyone knows?

glS
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heppoko_taroh
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1 Answers1

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Let $(e_n)$ be an orthonormal basis for a Hilbert space $H$ or any linearly independent sequence of unit vectors. Extend this sequence to a (Hamel) basis $(e_n) \cup (x_i)_{i \in I}$. Define$T(e_n)=ne_n$ and $T(x_i)=0$ for all $i$. This extends to an unbounded linear map from $H$ into $H$.

Kavi Rama Murthy
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  • Thank you for your answer. However, your answer is wrong, because $T$ is not defined. For example, if $x := \sum_{n=1}^\infty (1/n)e_n \in H$, then $Tx$ is not defined. – heppoko_taroh Jan 28 '21 at 09:37
  • Any map defined on basis elements extends uniquely to a linear map on the whole space. Your $x$ has another expression as finite linear combination of $e_n$'s and $x_i$'s and that expression is used to define $Tx$. @heppoko_taroh – Kavi Rama Murthy Jan 28 '21 at 09:44
  • Thank you for your answer. I was wrong and now I got it. – heppoko_taroh Jan 28 '21 at 10:03